One kilomole of an idea gas is compressed isothermally at 127 C from 1 to 10 atm

Question

One kilomole of an idea gas is compressed isothermally at 127 C from 1 to 10 atm in a piston-and-cylinder arrangement. Calculate the entropy change of gas, the entropy change of the surroundings, and the total entropy change resulting from the process, if the process is mechanically reversible and the surrounding consist of a heat reservoir at 127 C.

Please answer the question in step by step answers so that I will understand it. I will award points as soon as I get the answer. Thanks! Going to add 300 more points, need answer asap!

Explanation / Answer

We can use Tds equation to solve for entropy change. The equation derived for simple compressible system, which undergo internal reversible process.
Tds= du + pdv
integrating from T1 to T2, with gas ideal assumption, constant heat capacity

s2-s1 = Cv ln T2/T1 + R* ln P1/P2, isotermal T2= T1
= 0 + 8.314 . ln (1/10) = -19.14 kJ/kmol.K == > entropy change of gas

For a reversible isothermal change, W = -nRTln(1/10) so w = - 1 * 8.31 * 400 *ln(1/10) = 5704 KJ. As it's isothermal internal energy must be 0 so q = w. -> q= 7657.47 KJ

the entropy change of the surrounding =Qrev/T =7657.47Kj/400=19.14 KJ/Kmol.K

hence

the total entropy change resulting from the process = -19.14+19.14 =0

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