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 Cl_{2}

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.

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:

Wanted: ? grams Cl_{2}

Given:

V = 35.5 L

P = 100.0 kPa = 0.9872 atm

T = 100. ºC = 373.15 K

The Ideal Gas Law is:

We want to solve for moles, *n*, and multiply moles by the molar mass of Cl_{2}.

Rewriting the equation to solve for moles:

Plugging in our values for P, V and T:

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

The answer is 80.8 g Cl_{2}.

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 O_{2}

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:

Converting -73 ºC to K:

Wanted: ? g O_{2}

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:

To solve for moles, *n*:

Plugging in our values for P, V and T:

To determine the number of grams of O_{2}, we multiply the number of moles by oxygen’s molar mass.

*There are 32 g O _{2}.*

**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 CO_{2} gas?

Wanted: ? kPa

Given:

mass = 1.28 g CO_{2}

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 CO_{2} to moles, we need to divide by the molar mass of CO_{2}.

We also need to convert 35.1 °C to Kelvin:

Wanted: ? kPa

Given:

mass = 1.28 g CO_{2} = 0.0291 mol

V = 5.00 L

T = 35.1 °C = 308.25 K

The Ideal Gas Law is:

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

The pressure of CO_{2} is 0.147 atm.

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

*The pressure of CO _{2} in kPa is 14.9 kPa.*