## book question 9.13

Volume: $\Delta S = nR\ln \frac{V_{2}}{V_{1}}$
Temperature: $\Delta S = nC\ln \frac{T_{2}}{T_{1}}$

Lubna_Abdulmajeed_2A
Posts: 21
Joined: Wed Sep 21, 2016 2:59 pm

### book question 9.13

Question 9.13 in the textbook asks the following:
"During the test of an internal combustion engine, 3.00L of nitrogen gas at 18.5 degrees C was compressed suddenly (and irreversibly) to 0.500L by driving in a piston. In the process, the temperature of the gas increased to 28.1 degrees C. Assume ideal gas behavior. What is the change in entropy of the gas?"
So this is one of the questions where we have to calculate the change in entropy (nRln(v2/v1)) at constant initial temp. and then calculate the change in entropy = c*ln(t2/t1) at constant final volume and add them up to find the total entropy change. However, when calculating change in entropy from t1 to t2, the answer uses (delta)S=nRln(t2/t1), however if it is at constant volume shouldn't we calculate it using Cvm? so using the heat capacity at constant volume which would be 3/2(R) ?

Jessica Huang 1M
Posts: 26
Joined: Wed Sep 21, 2016 2:59 pm

### Re: book question 9.13

"In the process, the temperature of the gas increased to 28.1 degrees C."
During the process, volume decreased from 3.00 L to 0.500 L. Volume did not remain constant, so you cannot use Cv.
The solution works because entropy is a state function. You can calculate the change in entropy when volume change and then calculate the entropy when temperature changes.