Question

BEST RESPONSE GETS A FAN!

Suppose you had 1.0 mol samples of the following gases at STP. If the volume of each of the samples of gas was reduced to one tenth of its original size and the temperatures remained the same, which gas would have the lowest pressure?
A)C6H8
B)C3H4
C)N2
D)O2
E)Cl2

Answer 1

I'm pretty sure the only way you can approach this question is to assume we're talking about real gases. If we assume these compounds are behaving as ideal gases, the pressure would be the same for all of them even after volume is decreased.

For real gases, instead of using PV=nRT, you need to use:

\[(P+\frac{n^2a}{V^2})(V-nb)=nRT\]
Where a and b are constants whose value depends on the gas we're talking about. For real gases, the measured pressure is LOWER than what we'd expect for ideal gases, because the individual gas particles interact with each other (even weakly), lowering the number of collisions with the walls of the container that give rise to pressure. Therefore, a number is added (the one with the a) to correct for this.

On the other hand, for real gases, the measured volume is HIGHER than what we'd expect for ideal gases, because the gas molecules themselves have some volume (even if it's very small) that we also need to consider. Therefore, a number is subtracted (the one with the b) to correct for this.

So which gas will have the lowest pressure? We need to focus on the pressure term here. What will make gas particles interact with each other more strongly? Being larger - because then more van der Walls forces would be possible. How can we estimate the size? Look at the molar masses. The gas with the largest molar mass will probably have the lowest pressure.

C6H8 = 80 g/mol
C3H4 = 40 g/mol
N2 = 28 g/mol
O2 = 32 g/mol
Cl2 = 70 g/mol

I'd say A is your answer judging by the molar masses. Hope that helps!

Answer 2

Thank you, I said A but that was just an educated guess. Thanks for explaining it.