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.


2 Answers 2


Let's derive the appropriate realtionship...

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


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

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.


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

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.

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


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).

  • $\begingroup$ So in other words you are saying that the mixture of gases occupies the same volume (137 mL) as each individual gas? I thought gas volumes were additive. Are they not? $\endgroup$ Commented Oct 29, 2018 at 18:27
  • $\begingroup$ For formula 3 you need $P_1$ and $V_1$ of the dry gas to get the appropriate value for the product $P_1V_1$. Since there is water vapor in the gas you can correct either the pressure or the volume, but not both. Pressure is the easier thing to correct since you have to start there anyway. $\endgroup$
    – MaxW
    Commented Oct 29, 2018 at 18:37
  • $\begingroup$ In other words $P_1V_1 = 100147$. So it can either be $137\times731$ or $133\times753$. $\endgroup$
    – MaxW
    Commented Oct 29, 2018 at 18:46
  • $\begingroup$ See, that's where I get stuck. Why don't you need to correct both? As in, what is the physical reason? The way I picture it, you have two individual gasses, each of which makes an individual contribution to the total pressure, the total volume and the total moles. So to calculate the number of moles of the dry gas you would need to find the pressure of the dry gas (which was the first thing I did) and the volume of the dry gas (which was the second thing I did). Why is this wrong? Are the volumes not independent/ additive? Does each component have the same volume as the total volume? $\endgroup$ Commented Oct 29, 2018 at 19:13
  • $\begingroup$ $t$ = total, $w$ = water, $g$ = gas $$P_tV_t = P_wV_w + P_gV_g$$ $$P_gV_g = P_tV_t - P_wV_w = 753\times137 - 22\times137 = (753-22)\times137 = 100147$$ $\endgroup$
    – MaxW
    Commented Oct 29, 2018 at 19:21

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.

  • $\begingroup$ Since 1982 STP is 750.06 mm of mercury at 0 °C so the molar volume is 22.711 liters. $\endgroup$
    – MaxW
    Commented Oct 29, 2018 at 12:39
  • $\begingroup$ OK. Then 123 ml. $\endgroup$ Commented Oct 29, 2018 at 12:41

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