What's wrong with this gas law problem?

Question

I recently took a chemistry test in which the following problem was offered:

a 137 mL gas sample is collected over $\ce{H2O}$ at $753\ \mathrm{mmHg}$ and $22\ \mathrm{^\circ C}$. What is the volume of the dry gas at STP if the vapor pressure of $\ce{H2O}$ at $22\ \mathrm{^\circ C}$ is $22\ \mathrm{mmHg}$?

a) $106\ \mathrm{ml}$

b) $117\ \mathrm{ml}$

c) $123\ \mathrm{ml}$

d) $126\ \mathrm{ml}$

Here's my work:

To start with, I used Dalton's Law of partial pressures: $$p_\text{total}=p_\text{gas}+p_{\ce{H2O}}$$ $$p_\text{gas}=731\ \mathrm{mmHg}\ \left(\text{at}\ 22\ \mathrm{^\circ C}\right)$$

Then I used the following proportion to find the volume of the gas: $$\frac{p_\text{gas}}{p_\text{total}}=\frac{V_\text{gas}}{V_\text{total}}$$ $$V_\text{gas}=133\ \mathrm{ml}\ \left(\text{at}\ \mathrm{^\circ C}\right)$$

Which made sense to me because if the total number of moles of the gas mixture is not changing, and the temperature is not changing the ratio of the volumes should be the same as that of the pressures (shouldn't it?)

Finally I did a bit of dimensional analysis to change from $22\ \mathrm{^\circ C}$ to STP: $$133\ \mathrm{ml}\ \cdot\ \frac{731\ \mathrm{mmHg}}{760\ \mathrm{mmHg}}\ \cdot\ \frac{273\ \mathrm K}{295\ \mathrm K}$$ $$V_\text{gas}=118\ \mathrm{ml}\ \left(\text{at STP}\right)$$

I feel like I'm missing/forgetting a key concept here because each individual step made sense in my head (and still makes sense now). The answer the prof gave was d. This is bugging me a lot and I would appreciate any help.

Answer 1

Let's derive the appropriate realtionship...

$$PV = nRT\tag{1}$$

so

$$\dfrac{PV}{T} = nR\tag{2}$$

but $nR$ is constant for the given sample of gas, thus

$$\dfrac{P_1V_1}{T_1} = \dfrac{P_2V_2}{T_2}\tag{3}$$

For the sample the volume is 137 ml, the dry gas has a pressure of 753-22=731 mm Hg, and the temperature is 22 °C = 295 °K.

Now here is a bit of a wrinkle. What is STP? Until 1982 IUPAC defined STP as 760 mm of mercury at 0 °C = 273 °K. After 1982 IUPAC has defined STP as 750.06 mm of mercury at 0 °C = 273 °K.

For old IUPAC STP

$$V_2 = \dfrac{P_1V_1}{T_1}\times\dfrac{T_2}{P_2} = \dfrac{731\times137}{295}\times\dfrac{273}{760} = 121.95 = 122\text{ ml}$$

For current IUPAC STP

$$V_2 = \dfrac{P_1V_1}{T_1}\times\dfrac{T_2}{P_2} = \dfrac{731\times137}{295}\times\dfrac{273}{750} = 123.57 = 124 \text{ ml}$$

In looking at some possible variations, it seems the prof is using the wrong initial pressure, and the old IUPAC definition.

$$V_2 = \dfrac{P_1V_1}{T_1}\times\dfrac{T_2}{P_2} = \dfrac{753\times137}{295}\times\dfrac{273}{760} = 125.62 = 126\text{ ml}$$

NOTE

When collecting a gas over water you must subtract the partial pressure of water at the collection temperature from the atmospheric pressure, but you don't adjust the volume too.

Also this assumes that the atmospheric pressure given, 753 mm Hg, is for dry air. If you were measuring the pressure with a laboratory barometer then you'd have to adjust the measured laboratory pressure for the relative humidity (partial pressure of water).

Answer 2

The number of moles of gas collected is $$n=\frac{PV}{RT}=\frac{(0.137)(753/760)}{(0.08206)(295)}=0.00561\ moles$$

The mole fraction of dry air is $$x=\frac{731}{753}=0.9708$$So the number of moles of dry gas is (0.00561)(0.9708)=0.00544. At STP, 1 mole of dry gas occupies 22400 ml. So the volume of dry gas is (22400)(0.00544)=122 ml.