Real Gases

Real Gases

The Kinetic Molecular Theory describes the properties of ideal gases. Recall the tenets of the Kinetic Molecular Theory:

  1. A gas consists of a collection of small particles that travel in straight-line motion and obey Newton’s Laws.

  2. Gas particles occupy no volume.

  3. Collisions between particles are perfectly elastic, meaning that no energy is gained or lost during the collision.

  4. There are no attractive or repulsive forces between the particles.

 

The Kinetic Molecular Theory states that gas particles occupy no volume. But in reality, real gases DO occupy volume.

The Kinetic Molecular Theory says that collisions between particles are perfectly elastic, but real gases have inelastic collisions, which means that energy is lost when gas particles collide.

And finally, Kinetic Molecular Theory says that there are no attractive or repulsive forces between gas particles. Whereas, real gases DO attract and repel each other.

Let’s consider the volume of real gas particles.

Real Gases DO Have Volume

Molecules of real gases DO take up space and have volume, which reduces the effective volume of their container. This volume is reduced by nb, where n represents the number of moles of gas, and b is an experimentally determined constant specific to that gas. Therefore, the effective volume of a real gas is less than that of an ideal gas, because the volume of the particles themselves is accounted for:

V-nb

Where:

V = volume

n = moles of gas

b = an experimentally determined constant, specific to that gas.

The ideal gas law, taking into account this reduction in effective volume, can be rewritten as:

V-nb-real-gas

Real Gases Exert Pressure

When the effective volume of the container is reduced, the pressure inside the container is INCREASED. This pressure is increased by [latex]{\frac{{n^2a }}{{V^2 }}}[/latex],

where a is an experimentally determined constant specific to the gas.

P-real-gas

Where:

P = pressure

V = volume

n = moles of gas

a = an experimentally determined constant, specific to that gas.

Combining the effect of pressure and volume on a gas, the ideal gas law becomes Van der Waal’s Equation.

Van der Waal’s Equation

van-der-waals-2

Where:

P = pressure

V = volume

n = moles of gas

R = 0.081 L-atm-K-1-mol-1

a and b are experimentally determined constants specific to the gas

When does a Real Gas Behave Like an Ideal Gas?

A real gas behaves like an ideal gas when the gas particles stay as far away from each other as possible. This occurs at the conditions of low pressure and high temperature. Under low pressure and high temperature conditions, gas particles have little contact with each other thereby colliding less frequently and exerting minimal forces on each other.

Let’s look at a sample problem using Van der Waal’s equation:

Sample Problem

Using the Van der Waals equation, calculate the pressure in a 22.4 L vessel containing 1.00 mol of neon gas at 100 degrees Celsius (a = 0.211, b = 0.0171)

We are asked to find pressure, and we are given:

V = 22.4 L

n = 1.00 mol

T = 100 ºC

 

As with all gas problems we need to convert temperature into Kelvin:

373K

Van der Waal’s equation is:

van-der-waals-2

 

To solve for P, we first divide each side of the equation by V-nb:

pressure-1

 

Our equation becomes:

pressure-plus

 

Then, we subtract n2a/V2 from each side:

van-der-waals

 

Plugging in the given values, and solving for pressure, P:

137atm

 

 

Graham’s Law of Effusion of Gases

Graham’s Law of Effusion

Before we talk about Graham’s Law of Effusion, let’s explore what effusion is and distinguish it from diffusion.

Effusion is the movement of one type of gas particles through a hole.

effusion

 

Diffusion, on the other hand, is the movement of particles from areas of high concentration to areas of low concentration.

diffusion

 

Graham’s Law of Effusion talks about the relative rates of effusion as a function of the molar mass of gas particles. It can be applied equally well to the rates of diffusion of a gas, so sometimes the law is called Graham’s Law of Diffusion too. Graham’s Law of Effusion states that the rates of effusion of two gases are inversely proportional to the square roots of their molar masses at the same temperature and pressure:

grahams-law

 

In which:

M1 = molar mass of gas 1

M2 = molar mass of gas 2

Let’s look at a sample problem to learn how to use this equation.

Sample Problem

In an experiment, it takes an unknown gas 1.5 times longer to diffuse than the same amount of oxygen gas, O2. Find the molar mass of the unknown gas.

Let’s let M2 be the molar mass of the unknown gas since it’s in the numerator and therefore easier to solve for. That means that M1 is the molar mass of oxygen gas.

We know that the molecular formula of oxygen gas is O2, so its molar mass is 32.00 g/mol.

M2 = molar mass of unknown

M1 = 32.00 g/mol

The problem also says that the unknown gas takes 1.5 times longer to diffuse than oxygen. This means that oxygen diffuses 1.5 times faster, which means that the rate of diffusion of oxygen is 1.5 times as great. So, if the rate of diffusion of O2 is R (we are not given a magnitude, so we will use a variable in its stead), then the rate of diffusion of the unknown gas is 1.5 times as long or 1.5R.

Therefore, if the rate of diffusion of gas 2 is R, the rate of diffusion of gas 1 oxygen is 1.5R.

Rate of effusion of gas 1 = 1.5R

Rate of effusion of gas 2 = R

Substituting these values into our equation:

grahams-law

 

 

15R

The variable R cancels out on the left hand side of the equation:

15

 

To get M2 out of the square root, we square both sides:

152

 

1.5 raised to the second power is equal to 2.25:

225

 

Solving for M2:

72

 

The unknown gas has a molar mass of 72.00 g/mol.

 

Sample Problem

If it takes 20.0 minutes for 0.350 moles of H2S to effuse from a chamber, how long will it take for 0.175 moles of Kr to effuse from the same chamber?

The rate of effusion is expressed as moles per second, or moles per minute. Therefore, the rate of effusion of H2S is:

h2s-effusion

 

We want to know the length of time it will take for 0.175 mol Kr to effuse from the same chamber. Therefore, the rate of effusion of Krypton can be expressed as:

kr-effusion

 

The molar mass of H2S is 34.076 g/mol, and the molar mass of Kr is 83.80 g/mol.

molar-mass-h2s

 

molar-mass-kr

 

Our equation is:

grahams-law

 

We’ll let Kr be gas 2, so that time t will end up in the numerator:

rate-effusion-h2

 

0.0175 divided by 0.175 is equal to 0.100:

100t

 

And, solving the right-hand side of the equation:

100t157

 

Solving for t:

157min

 

It will take Kr 15.7 minutes to effuse from the same chamber.

 

Root Mean Square Velocity of a Gas

Root Mean Square Velocity

You already know that the velocity (or speed) of gas particles increases with increasing temperature. Another factor affecting the velocity of gas particles is their size — specifically, their molar mass. The following equation relates the speed of gas particles to both temperature and molar mass.

rms

 

where

[latex]\mu[/latex]rms represents root mean square velocity, which is a measure of the average speed of gas particles.

R equals Rydberg’s constant, which for the units in the above equation, equals:

R8314

 

T = temperature in Kelvin

M = molar mass in kg/mol (to match the units of Rydberg’s constant, R)

Let’s do a sample problem.

 

Sample Problem

What is the root mean square velocity of helium at 1 atm and 8.20 K?

The equation for Root Mean Square Velocity is:

rms

 

We are given a pressure of 1 atm and a temperature of 8.20 K. We can get the molar mass because we know the gas is helium. The molar mass of helium is 4.003 g/mol. We need to convert that to kg/mol.

 

mm-he

 

Plugging our values into the root mean square velocity equation:

226

 

The root mean square velocity of helium at 8.20 K is 226 m/s.

 

Sample Problem

What is the root mean square speed of nitrogen gas at STP?

Recall that STP stands for Standard Temperature and Pressure which is 273.15 K and 1 atm.

Nitrogen is a diatomic molecule, so the molecular formula for nitrogen gas is N2. Its molar mass is:

mm-n2

 

We need to convert the molar mass into the unit kg/mol:

mm-n2-kg

 

Plugging our values into the root mean square velocity equation:

493

 

The root mean square velocity of nitrogen gas at STP is 493 m/s.

Average Kinetic Energy of a Gas

Average Kinetic Energy of a Gas

The Kinetic Energy, also known as the energy of motion, of gas particles is directly proportional to temperature. The higher the temperature, the more kinetic energy the particles have. Another way of saying this is that, the higher the temperature, the faster the particles move.

This relationship between kinetic energy and temperature is described quantitatively through this equation:

ke-gas

 

This equation says that the Kinetic Energy (KE) per mole of particles is equal to three-halves times Rydberg’s constant times temperature.

Remember that temperature in all gas problems must be in the unit Kelvin.

Rydberg’s constant here is different than the one we have been using thus far, because the units are different. The Rydberg’s constant to use in this equation is:

R8314

 

This equation tells us that Kinetic Energy increases with temperature. But what about the relationship between KE and pressure? And between KE and volume?

Recall from Gay-Lussac’s Law that pressure is directly proportional to temperature:

Gay-lussac-small

 

As pressure increases, temperature increases, and vice versa.

Since pressure and temperature of a gas are directly proportional, and temperature and kinetic energy are directly proportional, we can surmise that pressure and kinetic energy are also directly proportional.

Kinetic Energy and Pressure are directly proportional.

What about Kinetic Energy and volume?

Recall from Charles’s Law that volume and temperature are directly proportional:

charles-law-small

 

As volume increases, so does temperature, and vice versa.

Since volume and temperature of a gas are directly proportional, and temperature and kinetic energy are directly proportional, volume and kinetic energy must also be directly proportional.

Kinetic Energy and Volume are directly proportional.

Let’s practice using our new equation in a sample problem.

Sample Problem

A 5.00 L flask is filled with a sample of fluorine at a temp of 301.8 ºC. Calculate the average kinetic energy.

Our equation relating KE and temperature is:

ke-gas

 

Remember that, for all gas problems, we need to convert degrees Celsius into Kelvin.

 

574K

 

Plugging in our value for temperature into the equation:

7170J

 

The average kinetic energy is 7170. J/mol.

Sample Problem

According to Kinetic Molecular Theory, if the temperature of a gas is raised from 100 °C to 200 °C, what will happen to the average kinetic energy?

Again, our equation relating kinetic energy and temperature is:

ke-gas

 

Remember, we need to convert temperature units to Kelvin.

373K

 

473K

 

To compare the kinetic energy of the two gases, we can look at the ratio of the kinetic energy of the second higher temperature of the gas to the lower first temperature:

127

 

Notice that both the fraction 3/2 and the constant R cancel out.

The kinetic energy of the gas increases by a factor of 1.27.

Kinetic Molecular Theory

Kinetic Molecular Theory

Kinetic Molecular Theory is a set of basic assumptions about gas particles that allows us to predict their behavior fairly accurately. These are the four tenets of Kinetic Molecular Theory (KMT):

  1. The particles in a gas are small and very far apart. Most of a gas’s volume is empty space.
  2. Gas particles move in straight lines, in constant random motion. As a result, particles collide with each other and with the walls of the container.
  3. Collisions between particles are perfectly elastic. No kinetic energy is lost.
  4. Gas particles do not attract or repel one another.

Partial Pressure and Mole Fraction

Partial Pressure and Mole Fraction

Mole fraction is the ratio of the number of moles of one component in a mixture to the total number of moles in the mixture. In the case of a mixture of gases, the mole fraction of one particular gas would equal the number of moles of that gas divided by the total number of moles in the mixture.

gas-mixture-mole-fraction

The mole fraction of gas A in the mixture above would be:

mole-fraction-A

 

where

X = mole fraction

n = number of moles

Mole fraction can be used to determine partial pressure of a gas in a mixture of gases. The partial pressure of a gas in a mixture of gases is equal to the total pressure of the mixture times that mole fraction of that particular gas:

partial-pressure-A

 

where

Pgas A = partial pressure of gas A

XA = mole fraction of gas A

Ptotal = total pressure of the mixture of gases

 

Let’s look at some sample problems.

 

Sample Problem

A mixture of helium (8.00 g) and argon (40.0 g) in a container at 300. K has a total gas pressure of 0.906 atmosphere. What is the partial pressure of helium in the mixture?

We are given the total gas pressure of 0.906. We simply need to multiply this total pressure by the mole fraction of helium to get the partial pressure of helium.

First, we need to convert our masses to moles:

mol-He

 

mol-Ar

 

The mole fraction of helium is:

mole-fraction-He

 

Notice that there are no units associated with mole fraction, because it is a ratio and the unit mol cancels out.

The partial pressure of helium is equal to the product of the mole fraction of helium and the total gas pressure:

partial-pressure-He

 

The partial pressure of helium in this mixture is 0.604 atm

Sample Problem

A mixture of 4.00 g of O2 and 6.00 g of CH4 is placed in a 20.0 L vessel at 0 °C. What is the partial pressure of each gas, and what is the total pressure in the vessel?

First, we can convert the 4.00 g of O2 and 6.00 g CH4 to their respective number of moles by dividing by their molar masses. We can then plug the total number of moles of gases into the Ideal Gas Law to get the total pressure. Once we have the total pressure, we can multiply that by the mole fraction of each gas to get the partial pressure of each gas.

Converting mass to moles:

mol-O2

 

mol-CH4

 

The total number of moles of gas in the mixture is the sum of the number of moles of these two gases:

 

499mol

 

We will plug this value for ntotal into the ideal gas law:

ideal-gas-law-small

 

And solve for P (which in this case is the total pressure because we we will substitute the total number of moles of gas for n).

ideal-gas-law-P

 

The problem provided us with the following additional variables:

Given:

V = 20.0 L

T = 0 °C

We need to convert temperature into Kelvin:

27315K

 

Now we plug our values into the above equation:

560atm

 

The total pressure of the mixture is 0.560 atm. To solve for the partial pressure of each gas, we will multiply this total pressure by the mole fraction of each gas. Recall that mole fraction of one gas in a mixture is equal to the number of moles of that gas in the mixture divided by the total number of moles of gas in the mixture:

mole-fraction-A

 

The mole fraction of O2 would be:

251

 

The mole fraction of CH4 is:

749

 

Notice that the sum of the mole fractions of each gas in the mixture equals 1.

We will now multiply the total pressure obtained above by the mole fraction of each gas to get the partial pressure of each gas.

partial-pressure-A

 

mole-fraction-O2

 

mole-fraction-CH4

 

The sum of the partial pressures of each gas in the mixture should be equal to the total pressure. 0.141 atm + 0.419 atm = 0.560 atm.

PO2 = 0.141 atm, PCH4 = .0419 atm

Dalton’s Law of Partial Pressure Collecting Gas over Water

Dalton’s Law of Partial Pressure allows us to determine the pressure (and from that, concentration) of a gas produced in a chemical reaction by collecting the gas over water.

collecting-gas-over-water

When gas is collected over water, the gas displaces the water in the flask as depicted above. The volume the gas displaces the water is the volume of the gas collected. If the flask is adjusted so that the water level inside the flask is equal to the water level outside of the flask, we know that the gas pressure inside the flask is equal to the gas pressure outside of the flask.

p-total

However, the gas in the flask is a combination of two gases really, the pressure of the gas PLUS the pressure of the water vapor, because water is in a continual state of vaporization and condensation.

p-atm

The value of the partial pressure of the water vapor can be found in a reference table, given that you know the temperature and atmospheric pressure. To determine the pressure of the gas, solve for Pgas:

p-gas

This experimental method only works for gases that are insoluble in water. If the gas were soluble in water then the amount of gas collected in the gaseous state would be less, as some of it would be dissolved in the water. In that case, the partial pressure of the gas collected would be less than the total amount of gas produced. Examples of gases that are soluble in water, and therefore cannot be collected in this way, are ammonia NH3 and hydrogen chloride HCl.

Let’s look at a sample problem.

Sample Problem

21.4 mL of hydrogen gas were collected over water at 15 ºC and 756.0 mm Hg. How many moles of hydrogen gas were collected? (vapor pressure of H2O at 15 ºC is 12.8 mm Hg).

From the volume and pressure of hydrogen gas, we can determine the number of moles of hydrogen gas collected using the ideal gas law. We know that the volume collected is 21.4 mL. To determine the pressure of gas collected, we must realize that the 756.00 mm Hg of gas collected over water is a mixture of hydrogen gas and water vapor. We’re given the partial pressure of water vapor as 12.8 mm Hg. We know that the total pressure in the flask is a sum of the partial pressure of hydrogen gas and the partial pressure of water vapor:

p-total-h2

Rewriting to solve for the partial pressure of hydrogen gas:

p-h2

Plugging in our values:

743

The pressure of the hydrogen gas collected is 743.2 mm Hg.

We want to now find the number of moles of hydrogen gas produced.

These are our givens:

P = 743.2 mm Hg

V = 21.4 mL

T = 15 ºC

And the ideal gas law is:

ideal-gas-law-small

 

We need to make sure the units of our variables match those of Rydberg’s constant R, L-atm-K-1-mol-1.

Pressure must be in the unit atmospheres, volume must be in the unit Liters and temperature must be in the unit Kelvin.

978atm

 

214L

 

288K

 

Rewriting the ideal gas law to solve for moles, n:

 

nPV

 

And plugging in our values for the variables P, V and T:

 

885mol

 

8.85 x 10-4 mol of H2 gas was collected.

Sample Problem

A 27.7 mL sample of CO2 was collected over water at 25.0 ºC and 1.00 atm. How many moles of CO2(g) were collected? (The vapor pressure of water at 25.0 ºC is 23.8 torr.)

Again, we are asked to solve for moles of gas. We will ultimately use the ideal gas law to do this. Keep in mind that a problem could go beyond the ideal gas law, and require us to use the calculated moles of a gas in a stoichiometric problem. This current problem does not go into stoichiometry.

We are given a total pressure of gas collected as 1.00 atm, and that the water vapor pressure is 23.8 torr. We know that the total pressure is the sum of the partial pressures of the gas and water vapor.

p-total-co2

Solving for the partial pressure of CO2:

p-co2

 

Since the total pressure is given in atm, and the vapor pressure of water is given in the unit torr, we need to first convert vapor pressure to atmospheres.

313atm

 

 

 

97atm2

 

We want to find mol CO2, n, and we’re given:

P = 0.97 atm

V = 27.7 mL

T = 25.0 ºC

We need to convert the unit of volume to Liter and the unit of temperature to Kelvin.

277L

 

298K

 

The Ideal Gas Law is:

ideal-gas-law-small

 

Rewriting to solve for n:

nPV

 

And, plugging in our variables for P, V and T:

 

11mol

 

0.0011 mol CO2 was collected.

 

 

Gas Stoichiometry

Gas Stoichiometry

The ideal gas law can be used to determine the number of moles of a gas given pressure, volume and temperature. The number of moles of a gas in a chemical equation can be used to determine the number of moles of any other species in the same reaction. Hence, we can use the ideal gas law to help solve stoichiometry problems.

Let’s consider a sample problem.

Sample Problem

Calculate the volume of chlorine gas at STP that is required to completely react with 3.50 g of silver, using the following equation:

2Ag(s) + Cl2(g) —> 2AgCl(s)

Wanted: ? L Cl2(g) at STP

Given: 3.50 g Ag

To convert 3.50 g Ag to L Cl2(g) at STP, we need to first convert grams Ag to mole Ag, mol Ag to mol Cl2. Once we have mol Cl2 we can plug it into the Ideal Gas Law and solve for volume. There is an easier way, however. Since this gas is at STP, we can use the conversion factor that 1 mol of gas at STP has a volume of 22.414 L. Recall that 22.414 L is referred to as the molar volume.

Conversion Factors:

Converting 3.50 g Ag to mol Ag: 1 mol Ag = 107.9 g Ag

Converting mole Ag to mole Cl2: 2 mol Ag = 1 mol Cl2

Converting mole Cl2 to L Cl2 at STP: 1 mol gas at STP = 22.414 L

363L

 

3.50 g Ag at STP requires 0.363 L of chlorine gas.

 

Sample Problem

Calcium carbonate decomposes to form carbon dioxide and calcium oxide:

CaCO3(s) —> CO2(g) + CaO(s)

How many grams of calcium carbonate will be needed to form 3.45 liters of carbon dioxide at 740 mm Hg and 121 °C?

Wanted: ? g CaCO3

Given:

V = 3.45 L CO2

P = 740 mm Hg

T = 121 °C

Since this gas is not at STP, we will need to use the ideal gas law to solve for moles of CO2. Once we have moles of CO2, we can use stoichiometry to determine the grams of CaCO3 that will be needed.

The ideal gas law is:

(1) ideal-gas-law-small

 

Rewriting the equation to solve for moles:

(2) nPV

 

Because Rydberg’s constant is in the units L-atm-K-1-mol-1, we need to make sure our values for P, V and T are in the same units.

Temperature must be converted to Kelvin:

394K

 

And pressure must be converted to atmospheres:

97atm

 

Rewriting our Given:

Given:

V = 3.45 L CO2

P = 740 mm Hg = 0.97 atm

T = 121 °C = 394.15 K

Plugging these values into Equation (2):

(2) nPV

 

10mol

 

Now that we know 0.10 mol CO2 are produced, we can determine how many grams of CaCO3 are needed to produce it?

Our conversion factors are:

Converting 0.10 mol CO2 to mol CaCO3: Our balanced chemical equation tells us that 1 mol of CO2 is produced from 1 mol CaCO3.

Converting mol CaCO3 to g CaCO3 requires the molar mass of CaCO3: 100.09 g CaCO3 = 1 mol CaCO3

Setting up our dimensional analysis problem:

10gcaco3

 

10. g of CaCO3 are needed.

Sample Problem

When chlorine is added to acetylene, 1, 1, 2, 2-tetrachloroethane is formed:

2Cl2(g) + C2H2(g) —> C2H2Cl4(g)

How many liters of chlorine will be needed to make 75.0 grams of C2H2Cl4 at 1.90 atm and 30 °C?

Wanted: ? L Cl2

Given: 75.0 g C2H2Cl4

P = 1.90 atm

T = 30 °C

Since we are given grams of C2H2Cl4, we will use that to determine the number of moles of chorine gas required to produce it. Once we have moles chlorine gas, we can use the ideal gas law to determine volume.

To convert 75.0 g C2H2Cl4 to mol Cl2, we need the following conversion factors:

Converting 75.0 g C2H2Cl4 to mol C2H2Cl4 requires the molar mass of C2H2Cl4: 1 mol C2H2Cl4 = 167.836 g C2H2Cl4

Converting mol C2H2Cl4 to mol Cl2 requires the mole ratio between the two species, obtained from the balanced chemical reaction: 1 mol C2H2Cl4 = 2 mol Cl2

894mol

 

0.894 mol Cl2 will make 75.0 g of C2H2Cl4. We will plug this value and the above pressure and temperature values into the ideal gas law to compute the volume of Cl2 gas.

But first, we need to convert temperature to Kelvin:

303K

 

Given:

n= 0.894 mol Cl2

P = 1.90 atm

T = 30 °C = 303.15 K

The ideal gas law (1) is:

(1) ideal-gas-law-small

 

Rewriting to solve for V:

ideal-gas-law-V3

 

Plugging in our values for n, T and P:

117L

 

11.7 L Cl2 gas are needed.

Ideal Gas Law and Density of a Gas

Ideal Gas Law and Density of a Gas

The Ideal Gas Law can also be used to determine the density of a gas. To derive that equation, we begin with the ideal gas law:

(1) ideal-gas-law-small

 

Density is defined as mass per unit volume.

(2)
density

 

Recall that we also derived the equation for the molar mass of a gas in the post Ideal Gas Law and Molar Mass of a Gas as:

(3) ideal-gas-law-mm

 

To solve for g, we multiply each side by PV/RT:

(4) PV4

 

The resulting equation is:

(5) ideal-gas-law-g

 

Plugging equation (5) into our equation for density (2):

(6)ideal-gas-law-density1

 

Notice that volume V cancels out:

(7) ideal-gas-law-density2

 

We now can use equation (7) to solve gas law problems for density.

 

Sample Problem

What is the density of SO2 gas at 1.18 atm and 26 °C?

The gas law equation (7) for density is:

(7)ideal-gas-law-density2

 

Wanted: ? Density

Given:

Molar mass of SO2 gas = 64.06 g/mol

P = 1.18 atm

T = 26 °C

We need to convert degrees Celsius to Kelvin:

299K

 

Plugging these values into equation (7):

308gl

 

The density of of SO2 gas is 3.08 g/L.

 

Ideal Gas Law and Molar Mass of a Gas

Ideal Gas Law and Molar Mass of a Gas

The ideal gas law can be used to determine the molar mass of a gas. Recall that molar mass has the units grams per mole. Given the volume, pressure and temperature of a gas, we can use the ideal gas law to determine the number of moles of that gas. If we are also given the mass of that gas, we can divide its mass by the number of moles to determine the grams per mole, which is equal to molar mass.

(1)molar-mass-gas

 

Rewriting this equation in terms of moles:

 

(2)  molesn

 

This confirms what we already know, that to convert mass to moles, we divide mass by its molar mass.

Recall the ideal gas law:

(3) ideal-gas-law-small

 

If we substitute equation (2) for moles (n) into equation (3), we get:

(4) PV

 

Rewriting:

(5) PV2

 

To solve for molar mass, we first cross-multiply:

(6) PV3

 

And divide each side by PV:

(7) ideal-gas-law-mm

 

Equation (7) can be used to solve for the molar mass of any gas, for which you know its mass, temperature, pressure and volume.

Sample Problem

A 0.276 g sample of gas occupies a volume of 0.270 L at 739 mm Hg and 98 °C. Calculate the molecular weight of this gas.

Molecular weight is the same thing as molar mass. We want to find the molecular weight of this gas:

Wanted: ? Molar mass

Given:

mass = 0.276 g

V = 0.270 L

T = 98 °C

P = 739 mm Hg

As with all gas law problems, we need to convert degrees Celsius to Kelvin:

371k

 

And all units must match the units of Rydberg’s constant L-atm-K-1-mol-1. So we must convert mm Hg to atm pressure.

972atm

 

Given:

mass = 0.276 g

V = 0.270 L

T = 98 °C = 371.15 K

P = 739 mm Hg = 0.972 atm

Plugging our values into Equation (7):

(7)ideal-gas-law-mm

 

32gmol

 

The molecular weight of this gas is 32.0 g/mol.

Sample Problem

Calculate the mass, in grams, of 3.50 L of NO gas measured at 35 °C and 835 mm Hg.

In this problem, we’re given the molar mass, or at least the identity of the gas, NO for which we can find its molar mass. And, we’re asked to determine the mass in grams.

To rewrite equation (7) to solve for grams, we multiply each side by PV/RT:

(8)PV4

 

The resulting equation is:

(9) ideal-gas-law-g

 

We will use this equation (9) to solve for mass in grams.

We need molar mass, P, V and T:

Given:

Molar mass of NO = 30.01 g/mol

V = 3.50 L

T = 35 °C

P = 835 mm Hg

Once again, we need to convert temperature to Kelvin, and pressure to atm.

308k

 

110a

 

Our updated given values are:

Given:

Molar mass of NO = 30.01 g/mol

V = 3.50 L

T = 35 °C = 308.15 K

P = 835 mm Hg = 1.10 atm

Plugging these values into equation (9)

457g

 

The mass of this sample of NO gas is 4.57 g.

The Ideal Gas Law and Mass

The ideal gas law can tell us the number of moles of a gas, given its volume, pressure and temperature. Once we know the number of moles, we can calculate the mass of a gas by multiplying the number of moles by molar mass.

Let’s do just that.

Sample Problem

How many grams of chlorine gas would occupy a volume of 35.5 L at a pressure of 100.0 kPa and a temperature of 100. ºC?

Wanted: ? grams Cl2

Given:

V = 35.5 L

P = 100.0 kPa

T = 100. ºC

Remember that, for all gas law problems, we need to convert degrees Celsius to Kelvin.

T373

Next, since Rydberg’s constant has the units L-atm-K-1-mol-1, pressure needs to be in the units atm.

Converting kPa to atm:

atm9872

Wanted: ? grams Cl2

Given:

V = 35.5 L

P = 100.0 kPa = 0.9872 atm

T = 100. ºC = 373.15 K

The Ideal Gas Law is:

ideal-gas-law

We want to solve for moles, n, and multiply moles by the molar mass of Cl2.

Rewriting the equation to solve for moles:

ideal-gas-law-n2

Plugging in our values for P, V and T:

ideal-gas-law-114

To determine the number of grams of chlorine gas, we multiply the number of moles by the molar mass of chlorine gas:

ideal-gas-law-808

The answer is 80.8 g Cl2.

Sample Problem

How many grams are in a sample of oxygen gas if the pressure is 1520 mm Hg, the volume is 8200 mL and the temperature is -73 ºC?

Wanted: ? g O2

Given:

P = 1520 mm Hg

V = 8200 mL

T = -73 ºC

We have a number of conversions to do before we can use the ideal gas law to calculate the number of moles of oxygen gas. Since the units of Rydberg’s constant are L-atm-K-1-mol-1, we need volume in Liters, pressure in atmospheres and temperature in Kelvin.

8200 mL is equal to 8.2 L (since there are 1000 mL in 1 L).

Converting 1520 mm Hg to atm, there are 760 mm Hg in 1 atm:

ideal-gas-law-2atm

Converting -73 ºC to K:

ideal-gas-law-200K

Wanted: ? g O2

Given:

P = 1520 mm Hg = 2.0 atm

V = 8200 mL = 8.2 L

T = -73 ºC = 200.15 K

The Ideal Gas Law is:

ideal-gas-law

To solve for moles, n:

ideal-gas-law-n2

Plugging in our values for P, V and T:

ideal-gas-law-1mol

To determine the number of grams of O2, we multiply the number of moles by oxygen’s molar mass.

ideal-gas-law-32g

There are 32 g O2.

Sample Problem

Dry ice is carbon dioxide in the solid state. 1.28 grams of dry ice are placed into a 5.00 L evacuated chamber that is maintained at 35.1 °C. What is the pressure in the chamber in kPa after all the dry ice has sublimed into CO2 gas?

Wanted: ? kPa

Given:

mass = 1.28 g CO2

V = 5.00 L

T = 35.1 °C

In this problem, we’re given mass in grams. We need to convert mass to moles, and then we can put that into the ideal gas law to calculate pressure.

To convert 1.28 g CO2 to moles, we need to divide by the molar mass of CO2.

ideal-gas-law-291mol

We also need to convert 35.1 °C to Kelvin:

ideal-gas-law-308K

Wanted: ? kPa

Given:

mass = 1.28 g CO2 = 0.0291 mol

V = 5.00 L

T = 35.1 °C = 308.25 K

The Ideal Gas Law is:

ideal-gas-law

To solve for pressure, we divide each side by V, and then, plugging in our values for n, V and T:

ideal-gas-law-147

The pressure of CO2 is 0.147 atm.

The problem asked for pressure in kPa, so we need to convert some units:

ideal-gas-law-149

The pressure of CO2 in kPa is 14.9 kPa.

 

Ideal Gas Law

Ideal Gas Law

One mole of ANY gas at standard temperature and pressure occupies a volume of 22.414 L.

Recall that standard temperature is 0°C or 273.15 K, and that standard pressure is 1 atm.

But what about gases that are NOT at STP? What volume do these gases take up? We can determine that using the ideal gas law.

The Ideal Gas Law is:

ideal-gas-law

P = pressure

V = volume

n = number of moles of gas

R = Rydberg’s constant = 0.0821 L-atm-K-1-mol-1

T = temperature (in Kelvin)

Sample Problem

How many moles of argon are there in a 22.4 L sample of gas at 2 atm and 0 °C?

Wanted: ? moles (n in the above equation)

Given:

V = 22.4 L

P = 2 atm

T = 0 °C

With all gas law problems, temperature needs to be in Kelvin. So we will first convert degrees Celsius to Kelvin.

273K

Given:

V = 22.4 L

P = 2 atm

T = 0 °C = 273.15 K

The Ideal Gas Law is:

ideal-gas-law

To solve for n, we’ll divide each side by RT:

ideal-gas-law-n

We get:

ideal-gas-law-n2

Plugging in our variables for P, V, R and T:

ideal-gas-law-2-mol

There are 2 moles of argon gas.

Sample Problem

1.00 mole of carbon dioxide at 1.00 atm and 0.00 °C occupies how much volume?

Wanted: ? V

Given:

n = 1.00 mol

P = 1.00 atm

T = 0.00 °C

Remember that temperature needs to be converted to Kelvin:

273K

The ideal gas law is:

ideal-gas-law

To solve for V, we divide each side by P:

ideal-gas-law-V

We get:

ideal-gas-law-v2

Plugging in our values for n, R, T and P:

ideal-gas-law-224

As expected, 1 mole of a gas at standard temperature and pressure occupies a volume of 22.4 L.

Avogadro’s Hypothesis

Avogadro’s Hypothesis

Avogadro’s Hypothesis says that equal volumes of gases at the same temperature and pressure contain the same number of moles.

In other words, a mole of oxygen molecules with a molar mass 32.00 g/mol and a mole of radon atoms with a molar mass of 222 g/mol occupy the same volume at equal temperature and pressure. But if the particles are larger, we would anticipate that 1 mole would take up a greater amount of space. Common sense would tell us that the bigger the molecule, the more space a mole of it would take up. This is true for solids and liquids; but not for gases.

You see, a gas is very different from a solid or a liquid. The molecules in a solid or liquid are very close together. I cannot put my hand through a piece of wood, but I can easily do so through air. This is because air is made up of mostly empty space, about 99.9% empty. On a molecular level, there’s a tremendous distance between the molecules compared to their size. If a gas is about 99.9% empty space, then the size of a molecule has really no effect on the volume of the gas.

There’s a certain number of particles that is very important in chemistry. That number is the mole. Which is approximately 6.02 x 1023 particles.

The volume that one mole of gaseous molecules occupies depends on:

Temperature and Pressure

Temperature. The higher the temperature, the greater the volume gas particles occupy. We know this from Charles’s Law.

Pressure. The higher the pressure, the smaller the volume gas particles occupy. We know this from Gay-Lussac’s Law.

So chemists established a standard temperature and pressure when they compare gases. They set standard temperature and pressure as 0 ºC and 1 atm, respectively.

The volume that 1 mole of any gaseous substance occupies at standard temperature and pressure, known as the molar volume, is 22.414 L.

Molar volume is a conversion factor we can use to solve stoichiometry problems involving gases at standard temperature and pressure (STP).

Warning: This conversion factor 22.414 L = 1 mol applies ONLY to gases, and ONLY at STP.

Sample Problem

How many moles are contained in 65.5 liters of CO2 gas at STP?

Since 1 mol of a gas at STP occupies 22.414 L, we would expect 65.5 L to occupy more than 1 mole.

Wanted: ? Moles

Given: 65.5 L of CO2 gas at STP

We want to go from 65.5 L of a gas at STP to moles. Our conversion factor is that, for any gas at STP, 1 mole = 22.414 L.

Conversion Factor: 1 mole of a gas at STP = 22.414 L

Setting up our problem:

292

 

65.5 L of CO2 gas at STP contain 2.92 moles.

Sample Problem

How many liters are occupied by 3.44 moles of CH4 gas at STP?

Wanted: ? liters

Given: 3.44 moles of CH4 gas at STP

Since we want to go from our given, 3.44 moles of a gas at STP, to liters, our conversion factor is that 1 mole of any gas at STP occupies a volume of 22.414 L.

Conversion Factor: 1 mole of a gas at STP = 22.414 L

Setting up our problem:

771

3.44 moles of CH4 gas at STP occupy 77.1 L.

Combined Gas Law

Combined Gas Law

The three gas laws Charles’s Law, Boyle’s Law and Gay-Lussac’s Law can be combined into one combined gas law relating pressure, volume and temperature of a gas. The combined gas law is:

combined-gas-law

We’ll learn how to use this equation by considering some sample problems.

Sample Problem

Carbon dioxide occupies a 2.54 L container at STP. What will be the volume when the pressure is 150 kPa and 26 ºC?

The volume 2.54 L is V1, and we are also told that the container is at STP, standard temperature and pressure. Standard temperature is 273.15 K, which is T1, and standard pressure is 1 atm or 101.3 kPa, which is P1. 150 kPa is P2 and T2 is 26 ºC. We want to find out the volume V2 when the pressure is 150 kPa and temperature is 26 ºC.

V1 = 2.54 L

T1 = 273.15 K

P1 = 1 atm = 101.3 kPa

V2 = ?

T2 = 26 ºC

P2 = 150 kPa

Before we plug these values into the Combined Gas Law, we need to convert all temperatures to Kelvin. To convert 26 ºC to Kelvin:

combined-299

Click here to learn more about converting between celsius and Kelvin temperature units.

The Combined Gas Law is:

combined-gas-law

To isolate the variable V2, multiply each side by T2/P2:

combined-gas-law-isolate

We get:

combined-gas-law-v2

Plugging in our values for P1, V1, T2, P2 and T1:

combined-gas-law-19L

The volume will be 1.9 L.

Sample Problem

The pressure of 8.40 L of nitrogen gas is decreased to one-half its original pressure, and its temperature is doubled. What is the new volume?

V1 = 8.40 L

When the pressure is reduced to one-half its original pressure, P2 = ½P1.

Since its temperature is doubled, T2 = 2T1.

We are asked to find the new volume, V2.

V1 = 8.40 L

T1 = T1

P1 = P1

V2 = ?

T2 = 2T1

P2 = ½P1

The combined gas law is:

combined-gas-law

To solve for V2, we multiply each side by T2/P2:

combined-V2

Our equation becomes:

combined-V22

Plugging in our values for P1, V1, T2, P2 and T1, we get:

combined-v2-prob

Notice that P1 and T1 cancel out:

combined-prob2

And we get:

combined-v2-336L

The new volume is 33.6 L.

The individual Gas Laws can be derived from the Combined Gas Law

Boyle’s Law

Charles’s Law

Gay-Lussac’s Law

See the next sample problem to find out how.

Sample Problem

Helium in a sealed syringe is compressed to a volume of 13 L. Its original volume was 21 L at 542 torr. Find the new pressure in torr.

The original volume of the gas is 21 L, so we’ll let that be known as V1. That makes 13 L V2. Since 542 torr is associated with V1 of 21 L, 542 torr is P1. We want to find the new pressure P2 in torr.

V1 = 21 L

P1 = 542 T

V2 = 13 L

P2 = ?

The combined gas law is:

combined-gas-law

Since temperature is not mentioned at all in this problem, we can assume it is constant before and after. Since temperature is constant, T1 = T2:

combined-boyles

Our equation becomes Boyle’s Law:

boyles-law

Solving for P2:

boyles-law-p2

Plugging in our values for P1, V1 and V2:

boyles-law-880t

The new pressure is 880 Torr.

Sample Problem

A sample of oxygen occupies a volume of 150. L at 89.0 °C. What will be the volume of oxygen when the temperature drops to 0.00 °C?

The original volume of the oxygen is 150. L which is V1. The original temperature T1 is 89.0 °C. The problem asks for the new volume of oxygen V2 when the temperature drops to the new temperature T2 = 0.00 °C.

V1 = 150. L

T1 = 89.0 °C

V2 = ?

T2 = 0.00 °C

First we need to convert temperature in celsius to temperature in Kelvin.

combined-gas-law-temp-conversions

V1 = 150. L

T1 = 89.0 °C = 362.15 K

V2 = ?

T2 = 0.00 °C = 273.15 K

The Combined Gas Law is:

combined-gas-law

Since there is no mention of pressure in this problem, it is assumed to be constant, the same before and after. Since P1 = P2:

combined-gas-law-charles

The combined gas law becomes Charles’s Law:

charles-law

Solving for V2, the equation becomes:

charles-law-v2

Plugging in our values for V1, T2, and T1:

combined-gas-law-113L

The new volume of oxygen is 113 L.

Standard Temperature and Pressure (STP)

Chemists define the standard temperature and pressure of gases, abbreviated STP, as 0 ºC and 1 atm pressure. Recall that 0 ºC is equal to 273.15 Kelvin.

Standard Pressure = 1 atm = 760 mm Hg = 101.3 kPa

Click here to learn how to convert among pressure units.

Click here to learn how to convert between degrees Celsius and Kelvin.

Temperature Conversion between Kelvin and Celsius

The unit of temperature for all gas problems is Kelvin. Whereas the celsius scale is based on the boiling point and freezing point of water, 0 ºC and 100 ºC respectively, the Kelvin scale is based on the lowest possible temperature absolute zero, which is 0 K. There are no negative numbers in the Kelvin scale.

kelvin-scale

To convert between celsius and Kelvin temperatures, use the following equation:

kelvin-conversion

The boiling point of water is 100 ºC. In Kelvin, the boiling point of water is:

kelvin-bp-water

The freezing point of water is 0 ºC. In Kelvin, the freezing point of water is:

kelvin-fp-water

Absolute zero is 0 K. In celsius, absolute zero is:

celsius-absolute-zero

Gay Lussac’s Law

Gay-Lussac’s Law

Gay-Lussac’s Law says that gas pressure is directly proportional to gas temperature when volume is held constant. The equation is:

gay-lussacs-law

Gay-Lussac’s law says that, as you increase the pressure of a gas, its temperature also increases, if volume is held constant. As you increase the temperature of a gas, its pressure also increases.

The following is a graph of gas pressure as a function of temperature. As you can see, pressure and temperature are directly related. As pressure is increased, so is temperature.

gay-lussac-graph

Sample problem:

A cylinder contains a gas with a pressure of 125 kPa at a temperature of 200. K. Find the temperature of the gas which has a pressure of 100. kPa.

The original pressure of the gas is 125 kPa so we’ll call that P1. The temperature at this pressure is 200. K, which we’ll call T1. The pressure on the gas is then changed to 100. kPa, which is called P2. We want to find the temperature at this new pressure, which is T2.

P1 = 125 kPa

T1 = 200. K

P2 = 100. KPa

T2= ?

Gay-Lussac’s Law is:

gay-lussacs-law

To solve for T2, we first cross-multiply:

gay-lussacs-cross-multiply

Which becomes:

gay-lussac-p1t2

Finally, to isolate T2, divide each side by P1:

gay-lussac-t2

To get:

gay-lussac-t2equals

Plugging in our values for T1, P2 and P1:

gay-lussac-160K

The temperature of the gas at this new pressure is 160. K.

Sample Problem

A container, designed to hold a pressure of 2.5 atm, is filled with 20.0 mL of air at room temperature (20 °C) and standard pressure (1 atm). Will it be safe to throw this container into a fire where temperatures of 600°C will be reached?

To determine whether it will be safe to throw this container into a fire of 600 °C, we need to determine whether the pressure will exceed 2.5 atm at 600 °C.

Therefore, we need to calculate the final pressure P2 and see if it is greater or less lthan 2.5 atm. This final pressure is the pressure when the temperature reaches 600 °C, which is T2. The initial temperature T1 is 20 °C and the initial pressure P1 is 1 atm.

T1 = 20 °C

P1 = 1 atm

T2 = 600 °C

P2 = ?

Remember that, for all gas law problems, we need to convert temperatures to K.

gay-lussacs-293

T1 = 20 °C = 293.15 K

P1 = 1 atm

T2 = 600 °C = 873.15 K

P2 = ?

Gay-Lussac’s Law is:

gay-lussac-p1t2

To isolate the variable P2, we need to multiply each side of the equation by T2:

gay-lussacs-isolate

We get:

gay-lussacs-p2

Plugging in our values for P1, T2 and T1:

gay-lussacs-3atm

Since P2 is 3 atm, which exceeds 2.5 atm, the container will explode under this pressure.

Charles’s Law

Charles’s Law

Charles’s Law says the gas volume is directly related to gas temperature. As the temperature of a gas increases, so does its volume. And as the volume of a gas increases, so does its temperature.

charles-law

 

V1 is the initial volume, T1 is the initial temperature and V2 is the new volume, and T2 is the new temperature.

Here are some examples of gas volume being directly related to gas temperature.

charles-law-balloon

The volume of the above balloon will expand as the flask is heated. The hot air balloon works by that same principle. Hot air balloons are inflated by firing up the air inside of them, thereby increasing their volume.

charles-law-hot-air-balloon

 

The following graph illustrates the direct relationship between volume and temperature as described by Charles’s Law.

charles-law-graph

 

As temperature increases, so does volume.

Let’s look at some sample problems.

Sample Problem

The volume of a gas at 25 ºC is 250 ml. Find its volume at standard temperature if pressure is held constant.

Since pressure is held constant, we’re not going to worry about it. Pressure will be the same before and after the temperature change. So what is the temperature change? The problem asks us to find its new volume at standard temperature. Standard temperature is actually a value. Standard temperature is equal to 0 ºC. And, since this is a gas law problem, and ALL gas law problems must be computed with the Kelvin temperature scale, we will use 273.15 K as standard temperature.

Charles’s Law is as follows:

charles-law

We are given a temperature of 25 ºC, which we’ll call T1, and a volume at that temperature of 250 mL, which we’ll call V1. We want to find out the new volume V2 when the temperature is changed to standard temperature, 273.15 K.

T1 = 25 ºC

T2 = 273.15 K

V1 = 250 mL

V2 = ?

Before we do anything, we need to convert all temperatures to the Kelvin scale. Therefore, we need to convert 20 ºC to K. We do this by adding to it 273.15.


T1298

T1 = 25 ºC = 298.15 K

T2 = 273.15 K

V1 = 250 mL

V2 = ?

To solve for V2 we need to isolate the variable. We’ll do that by first multiplying each side by T2.

charles-law-t2

T2 cancels out on the right hand side and our equation becomes:

charles-law-v2

Plugging in our values for V1, T2 and T1, we get:

v2229mL

The new volume is 229 mL.

Here’s another sample problem.

Sample Problem

5.00 L of a gas is collected at 100. K and then allowed to expand to 20.0 L. What is the new temperature in order to maintain the same pressure?

We are given an initial volume of 5.00 L, so that will be V1. The temperature at that initial volume is T1, 100. K. The gas is then allowed to expand to 20.0 L, which is V2.

V1 = 5.00 L

T1 = 100. K

V2 = 20.0 L

T2 = ?

Charles’s Law is:

charles-law

To solve for T2, we first need to get T2 out of the denominator. We do that by cross-multiplying.

charles-law-cross-multiply

We get:

V1T2

Finally, to isolate T2, divide each side by V1:

charles-law-next

And we get:

charles-law-t2t1v1

Plugging in our values for T1, V2 and V1, we get:

charles-law-400k

The new temperature required to maintain the same temperature is 400. K.

 

Boyle’s Law

Boyle’s Law

Boyle’s Law says that gas volume is inversely proportional to gas pressure when we hold temperature constant. Inversely proportional means that, as you increase one, the other decreases, and vice versa. As you increase the pressure of a gas, its volume decreases. And as you decrease the volume of a gas its pressure increases. The reverse is also true.

The equation for Boyle’s Law is:

boyles-law

This equation says that the original pressure of a gas P1 times its original volume V1 is equal to the new pressure P2 times the new volume V2.

This graph illustrates this inverse relationship. As volume decreases, pressure increases. At a volume of 12 L, the pressure of the gas is 1 atm. At the lower volume of 6 L, pressure as increased to 2 atm. When we reduced the volume to half its original value (from 12 L to 6 L), we doubled the amount of pressure (from 1 atm to 2 atm).

boyles-law-graph

Let’s solve some sample problems.

Sample Problem

A sample of hydrogen at 1.5 atm had its pressure decreased to 0.50 atm, producing a new volume of 750 mL. What was its original volume?

First, we need to identify our variables. We’re given a sample of hydrogen at 1.5 atm, so let’s label this value P1, the original pressure.

Then its pressure is decreased to 0.50 atm. We’ll label this new pressure P2.

P1 = 1.5 atm

P2 = 0.50 atm

When the pressure is decreased to 0.50 atm, we have a new volume of 750 mL, which we’ll label V2. We’re asked to find the original volume, which we’ll label V1.

V1 = ?

V2 = 1.5 atm = 750 mL

Boyle’s Law is:

boyles-law

To solve for our unknown V1, we isolate the variable by dividing each side of the equation by P1.

v1-isolate

Our equation now becomes:

V1

Plugging in our values for P2, V2 and P1:

250mL

The original volume was 250 mL.

Sample Problem

A 175 mL sample of neon had its pressure changed from 75 kPa to 150 kPa. What is its new volume?

We are given a volume of 175 mL, which we’ll call V1.

We are also told that the pressure changed from 75 kPa, which we’ll call P1, to 150 kPa, which we’ll call P2. We want to find its new volume, V2.

P1 = 75 kPa

P2 = 150 kPa

V1 = 175 mL

V2 = ?

boyles-law

To solve for V2, we isolate the variable by dividing each side of the equation by P2:

V2-isolate

Our equation now becomes:

V2

Plugging in our values for P1, V1 and P2:

88mL

The new volume V2 is 88 mL.

Dalton’s Law of Partial Pressure

Dalton’s Law of Partial Pressure says that the total pressure of a mixture of gases equals the sum of the pressures that each would exert if it was present alone. The equation for Dalton’s Law of Partial Pressure is: the total pressure of a mixture of gases is equal to the pressure of gas A if it were present alone, referred to as the partial pressure of gas A, plus the pressure of gas B if it were present alone, plus the pressure of gas C, etc., etc.

daltons-law

In the following illustration, we have gas A in a container by itself exerting a pressure PA. Gas B (the yellow particles) is in a container by itself, exerting a pressure of PB, and Gas C is in a container by itself (the blue particles) exerting a pressure of PC. If we put all three of these gases in the same container, then the total pressure of that container is going to be the sum of the individual partial pressures of each gas. The total pressure, then, is going to be the pressure of gas A, plus the pressure of gas B, plus the pressure of gas C.

daltons-law-image

 

Dalton’s Law of Partial Pressure flows from Kinetic Molecular Theory. Since gas particles are assumed to not take up any volume, each gas particle has access to the entire volume of the container as if it were alone, and no other gas were present.

Let’s work through some sample problems.

Sample Problem

A mixture contains carbon dioxide with a partial pressure of 125 mm Hg, and oxygen with a partial pressure of 275 mm Hg. What is the total pressure of the mixture?

We know that the total pressure of the mixture is the sum of the individual partial pressures of the gases, according to Dalton’s Law of Partial Pressure:

400mmHg

The total pressure of the mixture is 400. mm Hg.

Sample Problem

A mixture of argon and neon has a total pressure of 1.50 atm. If the partial pressure of neon is 1.25 atm, what is the partial pressure of argon?

daltons-law

In this example, we are given the total pressure as 1.50 atm, and the partial pressure of neon, 1.25 atm. The total pressure 1.50 atm is the sum of the partial pressures of the two gases, neon 1.25 atm, and the partial pressure of argon.

ptotal

Therefore, the partial pressure of argon is equal to the total pressure minus the partial pressure of neon.

25atm

The partial pressure of argon is 025 atm.

 

 

Kinetic Molecular Theory of Gases

Kinetic Molecular Theory of Gases

The behavior of gases that we study in basic chemistry is based on the following four assumptions, known collectively as the Kinetic Molecular Theory of motion. Making these assumptions allows us to predict properties of gases that hold true most of the time. They are:

  1. A gas consists of a collection of small particles that travel in straight-line motion and obey Newton’s Laws.
  2. Gas particles occupy no volume.
  3. Collisions between particles are perfectly elastic, meaning that no energy is gained or lost during the collision.
  4. There are no attractive or repulsive forces between the particles.

How to Convert Units of Gas Pressure

Pressure is measured in many different units, including millimeters mercury (mm Hg), atmospheres (atm), Torr, Pascals , pounds per square inch (psi) and Bar. The conversion factors among these units are as follows:

pressure-unit-conversions

A Torr is equal to mm Hg, named in order of Evangelista Torricelli, the creator of the first barometer.

Let’s do some sample problems converting among units.

Sample Problem

The weather news gives the atmospheric pressure as 1.07 atm. What is this atmospheric pressure in mm Hg?

Wanted: ? mm Hg

Given: 1.07 atm

To convert from atmospheres (atm) to mm Hg, we need the conversion factor, 1 atm = 760 mm Hg.

Conversion Factor: 1 atm = 760 mm Hg

813mmHg

1.07 atm is equivalent to 813 mm Hg. Notice that the answer is in the same number of significant figures as the given value of 1.07 atm.

Sample Problem

The atmospheric pressure in a certain location is 761.3 mm Hg. What is this pressure in Pa?

Wanted: ? Pa (pascals)

Given: 761.3 mm Hg

Conversion Factor: We need to go from mm Hg to Pa, so we use the conversion factor 760 mm Hg = 101325 Pa.

1015Pa

761.3 mm Hg is equivalent to 1.015 x 105 Pa.

Sample Problem

The air pressure inside a submarine is 0.62 atm. What would be the height of a column of mercury balanced by this pressure?

The height of a column of mercury is measured in mm Hg (millimeters is a measure of height). Therefore, we need to convert pressure from atm to mm Hg.

Wanted: ? mm Hg

Given: 0.62 atm

Conversion Factor: 1 atm = 760 mm Hg

470mmHg

0.62 atm is equivalent to 470 mm Hg.

Sample Problem

Determine the gas pressure in atm in the open-end manometer below.

manometer

Recall that, to read a manometer, when the side open to the air is lower than the side open to the gas, the air pressure exceeds the gas pressure. The difference in height between the two columns tells the difference in pressure between the two.

For more help on how to read a manometer, click here.

Since the air pressure is 0.960 atm, and the air pressure is greater than the gas pressure by 45 mm Hg, the gas pressure will be 0.960 atm – 45 mm Hg. Since the problem asks for the gas pressure in units of atmospheres, we need to first convert 45 mm Hg to atm.

Wanted: ? atm

Given: 45 mm Hg

Conversion Factor: 1 atm = 760 mm Hg

59atm

Now that we’ve converted 45 mm Hg to 0.059 atm, we can finish answering the problem.

Gas Pressure = Air Pressure – 0.059 atm.

Gas Pressure = 0.960 atm – 0.059 atm = 0.901 atm.

The pressure of the gas is 0.901 atm.

Reading a Barometer and Manometer

How to Read a Barometer

The barometer was developed by Evangelista Torricelli in 1643. It measures atmospheric pressure.

barometer

Atmospheric pressure pushes down on the mercury in the basin, causing the column to rise. The height of the column is equal to the Atmospheric Pressure. Atmospheric Pressure, then, is measured by measuring the height of the column, and therefore it’s often given in millimeters mercury (Hg).

Pressure can also be measure with a manometer, which is U-shaped and measures pressure difference. There are two types of manometers, open-end and closed-end.

How to Read a Manometer

Open-End Manometer

The following is an open-end manometer.

manometer2

The liquid inside is usually mercury. When both sides are open to the air, the level of the liquid will be the same on both sides. When a gas of pressure Po is connected to one side, leaving the other side open to the air with a pressure Pa as illustrated above, the pressure of the gas can be determined by taking the difference in heights between the two columns. Since the mercury column on the side open to the air with pressure Pa is lower than the column on the side open to the gas, the air pressure is greater than the gas pressure. The gas pressure Po then, is Pa minus H.

Closed-End Manometer

A closed-end manometer is not open to the air. To read a closed-end manometer, take the difference between the heights of the two columns of mercury.

Sample Problem

In a closed-end manometer, the mercury level was 690. mm higher on the closed end than on the gas side. What was the pressure of the gas in mm Hg?

Since this problem involves a closed-end manometer, the pressure of gas is equal to the difference in height of the two columns. Since that difference in height is 690. mm Hg, the gas pressure equals 690. mm Hg.

Sample Problem

In an open end manometer, atmospheric pressure was 760. mm Hg, and the mercury level was 120. mm higher on the right side than the left. What was the gas pressure? Atmospheric pressure = 760. mm Hg.

manometer120

Since the mercury column on the side open to the gas is lower than that on the side open to the air, we know that the gas pressure is greater than the atmospheric pressure. The difference in pressure between the gas and air is equal to the difference in height between the two columns, 120. mm Hg.

Therefore, the gas pressure is greater than the air pressure by 120. mm Hg.

Gas pressure = 760. mm Hg + 120. mm Hg = 880. mm Hg

The pressure of the gas is 880. mm Hg.

Sample Problem

Assuming that the valve is open, what pressure, in mm Hg, is the gas exerting?

1013kPa

This is an open-end manometer, because one side is open to the air. Since the mercury column on the side open to the air is lower than that of the column open to the gas, we know that the air pressure is greater than the gas pressure. The difference in these two pressures is equal to the difference in height between the two columns, 40 mm Hg.

The gas pressure is less than the air pressure by a difference of 40 mm Hg.

Therefore the gas pressure is:

Gas pressure = 760 mm Hg – 40 mm Hg = 720 mm Hg

The pressure the gas is exerting is 720 mm Hg.

 

Images:

Mercury Barometer courtesy of Wikipedia http://en.wikipedia.org/wiki/File:MercuryBarometer.svg

Mercury Manometer courtesy of Wikipedia http://en.wikipedia.org/wiki/File:Utube.PNG

Two manometers side by side courtesy of images.yourdictionary.com

Gas Pressure

The concentration of gases is most often expressed as Gas Pressure. Pressure is defined as force per unit area. The equation for pressure is:

Pressure-defn

P = pressure

F = force

A = area

Gas Pressure is the force gases exert on their container per unit area.

This equation can be rewritten to solve for either Force or Area as well.

F=pxa

A=FP

Atmospheric pressure is a specific type of gas pressure referring to the pressure the atmosphere exerts on us.

At higher altitudes atmospheric pressure decreases.

Let’s consider the relationship among these three variables, pressure, force and area. Nails are easier than screws to hammer into wood because the area of contact between the nail and the wood is reduced to a small point. The force the hammer exerts on the nail is applied to an extremely small area, at the point of contact, which allows the hammer to apply a great amount of pressure. If a screw were hammered into a piece of wood with an identical force, the wider point of contact between the base of the screw and the piece of wood means that the screw exerts a smaller force.

hammer-screw

Looking at these two rectangular prisms below, we notice that the same force of gravity is acting upon them, because they each have the same weight.

pressure-blocks-area

However, the pressure each exerts is different.

Pressure-defn

The blocks on the right exert more pressure because the same force is being applied over a smaller area.

Sample Problem. 

If a rock exerts a force of 25 newtons over an area of 5.00 m2, how much pressure is the rock putting on the ground?

pressure-5

The rock exerts a pressure on the ground of 5.0 N/m2.

Sample Problem.

The pressure of a nail was measured at 350 Pascals (Pa). What force is exerted by the nail if the surface area is 0.17 m2?

Wanted:

Force = ?

Given:

P = 350 Pa

A = 0.17 m2

We rewrite the equation to solve for force:

pressure-60

A nail exerting a pressure of 350 Pa over an area of 0.17 m2 exerts a force of 60. Pam2.

Sample Problem.

What area is covered by a box that exerts 40 newtons of force and a pressure of 20 Pa?

Wanted:

Area = ?

Given:

F = 40 N

P = 20 Pa

We rewrite the equation to solve for area:

pressure-2

The box covers an area of 2 N/Pa.