# Finding the increase in entropy of the universe in gas expansion

Gold Member
Homework Statement:
Two boxes of equal volume ##V_0## are separated by a membrane. The first one to the left, thermally isolated contains ##n## moles of perfect biatomic gas at pressure ##p_i## and temperature ##T_i##. The second box is empty, and in contact with a source of heat at a temperature which is ##T_i/2##. At a certain instant the membrane is removed and the gas reaches a new equilibrium. Find: a) the final pressure of the gas ##P_f## b) the heat ##Q## exchanged with the source of heat c) the variation of entropy of the universe ##\Delta U_{univ}##
Relevant Equations:
##PV=nRT, \Delta U=nC_V \Delta T=Q-L, \Delta S=nC_V \ln(\frac{T_f}{T_i})+nR\ln(\frac{V_f}{V_i})##
a) ##P_f=\frac{nRT_f}{V_f}=\frac{nR\frac{T_i}{2}}{2V_0}=\frac{1}{4}\frac{nRT_i}{V_0}=\frac{1}{4}P_i##

b) ##Q=\Delta U=nC_V \Delta T=n\frac{5}{2}R(-\frac{T_i}{2})=-\frac{5}{4}nRT_i=-\frac{5}{4}P_i V_0## (##L=0## since the gas expands in a vacuum;

Now, (a) and (b) are both correct but not (c), for which I get:

c) ##\Delta S_{system}=nC_V \ln(\frac{T_f}{T_i})+nR\ln(\frac{V_f}{V_i})=n\frac{5}{2}R\ln(\frac{\frac{T_i}{2}}{T_i})+nR\ln(\frac{2V_0}{V_0})=-\frac{3}{2}nR\ln(2)## I understand that the entropy of the system decreases since heat goes out of the system so I guessed that the entropy of the universe should go up by the same amount but apparently this is wrong as in the solution given in the text is ##n\frac{5}{2}R-\frac{3}{2}nR\ln(2)##. What is it that I am missing in finding the entropy of the universe? How should I reason about such a problem in general? Thanks

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Mentor
The surroundings is being treated as an ideal constant temperature reservoir at temperature ##T_i/2##. The amount of heat it receives at this temperature is ##+\frac{5}{4}nRT_i##. So what is its entropy change?

Otherwise, you did a great job of analyzing this.

Gold Member
The surroundings is being treated as an ideal constant temperature reservoir at temperature ##T_i/2##. The amount of heat it receives at this temperature is ##+\frac{5}{4}nRT_i##. So what is its entropy change?
It is ##\frac{Q}{T_i /2}=\frac{\frac{5}{4}nRT_i}{\frac{T_i}{2}}=\frac{2}{T_i}\frac{5}{4}nRT_i=\frac{5}{2}nR## so ##\Delta S_{universe}=\frac{5}{2}nR-\frac{3}{2}nR\ln(2)=\frac{nR}{2}(5-3\ln(2))##. So the entropy change of the universe is the sum of the entropy change of the system + entropy change of its surroundings, right?

Chestermiller
Mentor
It is ##\frac{Q}{T_i /2}=\frac{\frac{5}{4}nRT_i}{\frac{T_i}{2}}=\frac{2}{T_i}\frac{5}{4}nRT_i=\frac{5}{2}nR## so ##\Delta S_{universe}=\frac{5}{2}nR-\frac{3}{2}nR\ln(2)=\frac{nR}{2}(5-3\ln(2))##. So the entropy change of the universe is the sum of the entropy change of the system + entropy change of its surroundings, right?
Yes. Very nice.

Gold Member
Yes. Very nice.
You have been very helpful: thanks!