Calculate Entropy Change for Adiabatic Process of Argon Gas

[bullados]

For quasistatic processes, the entropy of an α-ideal gas changes in the following way:

ΔS = nR ln(Vf/Vi) + αnR ln(Tf/Ti)

as discussed in lecture and in Wolfe's Notes. In this problem you will compute the contributions to S from the V and T terms separately, then add them up to find the total entropy change for an adiabatic process.

Argon gas, initially at pressure 100 kPa and temperature 300 K, is allowed to expand adiabatically from 0.01 m^3 to 0.029 m^3 while doing work on a piston.

a) What is the change in entropy due to the volume change alone, ignoring any effects of changing internal energy?

I have tried every combination of terms that I can think of within PV=nRT to find nR. Here are the three expressions that I have come up with, none of which are correct based on the answer to this question...

nR = 100 * 0.01e-3 / 300 (the expression with kPa as the units of pressure)

nR = 100e3 * 0.01e-3 / 300 (the expression with Pa as the units of pressure)

nR = (100e3 / 1.013e5) * 0.01e-3 / 300 (the expression with atm as the units of pressure)

Help, please! I'm not sure what I'm screwing up on! Thanks to anybody who can help with this...

 

[OlderDan]

I have tried every combination of terms that I can think of within PV=nRT to find nR. Here are the three expressions that I have come up with, none of which are correct based on the answer to this question...

nR = 100 * 0.01e-3 / 300 (the expression with kPa as the units of pressure)

nR = 100e3 * 0.01e-3 / 300 (the expression with Pa as the units of pressure)

nR = (100e3 / 1.013e5) * 0.01e-3 / 300 (the expression with atm as the units of pressure)

Help, please! I'm not sure what I'm screwing up on! Thanks to anybody who can help with this...

I think it will help you if you write the units with the numbers to make sure you have the result exressed in the units you need. I assume you want nR in units of Joules/K. I think I see an extraneous e-3 in the first two, and I have no idea why you tried the third one.

 

[bullados]

The e-3 is the conversion factor from m^3 to liters, which is why it's included. The third is because I've had success going from Pascals to Atmospheres before with this formula, and I'm a little n00bish with using it. Lemme re-write those with units...

nR = 100 kPa * 0.01 m^3 * (1 L / 1e-3 m^3) / 300 K
nR = 100e3 Pa * 0.01 m^3 * (1 L / 1e-3 m^3) / 300 K
nR = (100e3 Pa * (1 atm / 1.013e5 pa) ) * 0.01 m^3 * (1 L / 1e-3 m^3) / 300 K

I think I just found my mistake. I feel like an idiot for that one...