# Ideal Gases

## Problems 3

Summary Problems 3

Problem : What is the density of carbon dioxide at 298 K and 1.0 atm?

Use the equation for gas density:

 d = = μ = 44.0 . If we use the value of R0.0821 we can stick with the original units. Plugging in these values, we find that the density d of CO2 is 1.8 g/L.

Problem : 10 L of an unknown gas has a mass of 10.8 grams at a temperature of 310 K and 1.2 atm. What is the molar mass of the gas?

Rearrange the gas density equation above to solve for μ:

 μ = If we use R = 0.0821 we can avoid converting units. The density d is the mass of the gas divided by the volume. Plugging in values, we find that the molar mass μ is 23 grams per mole.

Problem : A syringe of 0.010 mol O2 and a syringe of 0.060 mol H2 are forcibly injected into an evacuated glass jar of volume 1.0 L and temperature 273 K. What is the total pressure, and what are the partial pressures, of O2 and H2?

There are 0.010 + 0.060 = 0.070 mol of gas in the jar. Since we know volume and temperature, we can rearrange PV = nRT to find the total pressure. Using the value R = 0.0821 , we find that Ptot = 1.6 atm.

To solve for the partial pressures of each gas, we use Charles' law. The mole fraction of O2 is 0.010/0.070 = 0.14, so the partial pressure of O2 is 14% of the total pressure, or 0.23 atm. Likewise, the mole fraction of H2 is 0.060/0.070 = 0.86, so the partial pressure of H2 is 1.4 atm.

Alternatively, you could have found the partial pressures directly and summed them to find the total pressure. Try both ways to see which works best for you.

Problem : The volumes of containers A and B are kept separate by a stopcock. Both containers have a volume of 10-3 m3 and are at a temperature of 273 K. The gas a in container A is at a pressure of 2.00×105 Pa. The gas b in container B is at a pressure of 1.00×105 Pa. What are the partial pressures of a and b after the stopcock has opened and the system has equilibrated?

You could approach this problem by calculating the number of moles of gas using PV = nRT, then resolving for the pressures. However, you can take a tremendous shortcut if you remember Dalton's law: the pressure a gas exerts in a mixture is the same as the pressure it would exert if alone. Since we know the final and initial volumes of the system and the initial pressure of each gas, we can use Boyle's law to calculate the final pressure contribution of each gas. For gas a, rearrange Boyle's law to solve for the final pressure Pa2:

 Pa2 = = = = 1.0×105 Pa = 1/2 for both gases. A similar calculation for gas B gives Pb2 = 0.50×105 Pa.