## Question regarding 'C' value in Entropy calculations

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

Sabrina Smelser 3D
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Joined: Fri Sep 26, 2014 2:02 pm

### Question regarding 'C' value in Entropy calculations

A practice problem was stated as such:
Suppose 150.0g of ethanol at 22.0 degrees Celsius is mixed with 200.0g of ethanol at 56.0 degrees Celsius at constant atmospheric pressure in a thermally insulated vessel. If the final temperature after mixing is 300.90K, calculate the change in entropy for hot ethanol, cold ethanol, and the total change in entropies combined.

A solutions manual used specific heat capacity for the C value in the entropy equation, when I thought we were to use the Pressure constant (5/2R). Can someone please explain to me why this is?

martha-1I
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### Re: Question regarding 'C' value in Entropy calculations

The pressure and volume constants for heat capacity are only used when the gas in context is an ideal gas or a noble gas. If it is stated that the ideal gas is at constant volume then we use 3/2R and if it is stated that the ideal gas is at constant pressure then we use 5/2R.

Justin Le 2I
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### Re: Question regarding 'C' value in Entropy calculations

What Martha said about heat capacity is specifically for monatomic gases.

martha-1I
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### Re: Question regarding 'C' value in Entropy calculations

Yes. Thank you for making that clear, Justin. What I described is specifically for monoatomic gases.

Quynh Nguyen 2O
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Joined: Fri Sep 26, 2014 2:02 pm

### Re: Question regarding 'C' value in Entropy calculations

This is a question in one of our practice midterms (midterm 2010 Q4A.)

During the test of an internal combustion engine, 3.00 L of nitrogen gas at 18.5 C was compressed suddenly ( and irreversibly) to .500 L by driving in a piston. In the process the temperature of the gas increased to 28.1 C. Assume ideal behavior and 1.00 mile of nitrogen gas. What is the change in entropy of the gas?

They used (3/2) R which is constant volume. But I would have used (5/2) R because the pressure is constant not volume in this problem, so I don't understand why they used constant volume.

martha-1I
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### Re: Question regarding 'C' value in Entropy calculations

I think the reason for why they used constant volume is because they compressed the engine which caused the pressure to change but the volume from that point on to stay the same. Since it is not stated that it decompressed or compressed more, we can assume the volume stayed the same after it had been compressed suddenly and that the reaction began after the sudden compression.

Chem_Mod
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### Re: Question regarding 'C' value in Entropy calculations

Ethanol is a liquid, and as such, neither 3/2R nor 5/2R can be used, since those are only for ideal, monatomic gases. Ethanol has its own experimentally determined heat capacities.