A 425-mL sample of hydrogen is collected above water at 35°C and 763 torr. Find the volume of the hydrogen sample when

Question

A 425-mL sample of hydrogen is collected above water at 35°C and 763 torr. Find the volume of the hydrogen sample when the temperature falls to 23°C, assuming the barometric pressure does not change. (vapor pressures of water : at 35°C, 42.2 torr ; at 23°C, 21.1 torr).

Can someone please tell me if my reasoning is correct? (or steps on how to get to the answer if I am doing this wrong)

What I did:

At 35°C = 308K:

P_H2 = (763/760)atm - (42.2/760)atm

P_H2 = 0.948421053 atm

At 23°C = 296K

P_H2 = (763/760)atm - (21.1/760)atm

P_H2 = 0.976184211 atm

Then I used the equation: (P₁V₁) / (n₁T₁) = (P₂V₂) / (n₂T₂)

Since the moles seem to be constant (as they do not seem to be changed in any way) I reduced this equation to: (P₁V₁) / T₁ = (P₂V₂) / T₂

I plugged in 0.948421053 atm for P1, 0.425L for V1, and 308K for T1. I then plugged 0.976184211 atm for P2, and 296K for T2. I then rearranged the equation to isolate V2. I got 0.3968L which is about 397 mL (when rounding up).

The answer in the book says 396 mL. Is the answer in the book just rounded down? Or is there something wrong in my work that is making it closer to 397mL than 396 mL?

Answer 1

Your approach seems to be fine, although you could have saved yourself a little math be leaving the pressure in mm Hg and not converting to atmospheres. That conversion was not necessary. As far as the answers go, you should round your calculations to 3 significant figures, so I'm assuming that the answer in the book is rounded differently than what you did. Bottom line ... your approach is spot on and it's good that you knew to subtract the vapor pressure of water from the vapor pressure of the gas.