Ideal Gases

Use the equation for gas density:

μ = 44.0  . If we use the value of R 0.0821  we can stick with the original units. Plugging in these values, we find that the density d of CO ₂ is 1.8 g/L.

Rearrange the gas density equation above to solve for 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.

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 P _tot = 1.6 atm.

To solve for the partial pressures of each gas, we use Charles' law. The mole fraction of O ₂ is 0.010/0.070 = 0.14, so the partial pressure of O ₂ is 14% of the total pressure, or 0.23 atm. Likewise, the mole fraction of H ₂ is 0.060/0.070 = 0.86, so the partial pressure of H ₂ 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.

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 P _a2 :

= 1/2 for both gases. A similar calculation for gas B gives P _b2 = 0.50×10⁵ Pa.