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 CO_{2} 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 O_{2} and a syringe of 0.060 mol H_{2} 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 O_{2} and
H_{2}?

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.

Problem :

The volumes of containers A and B are kept separate by a stopcock. Both
containers have a volume of
10^{-3} m^{3} and are at a temperature of 273 K. The gas a in container A is
at a pressure of 2.00×10^{5} Pa. The gas b in container B is at a pressure of 1.00×10^{5} 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 P_{a2}:

P_{a2}

=

=

=

=

1.0×10^{5} Pa

= 1/2 for both gases. A similar calculation for gas B gives
P_{b2} = 0.50×10^{5} Pa.