If 2.07 moles of H2(g) (cp = 28.8 J k-1 mole-1) is compressed reversibly and adi

Question

If 2.07 moles of H2(g) (cp = 28.8 J k-1 mole-1) is compressed reversibly and adiabatically form an initial pressure of 3.00 atm at 400 K to a final pressure of 7.00 atm, calculate Delta S for the gas. Suppose a different path is used. The gas is first heated at constant pressure to the same final temperature, and then compressed reversibly and isothermally to the same final pressure. Calculate Delta S for the two segments of this path, and the total entropy change for the process. Is the total entropy change different for the two paths?

Explanation / Answer

P1 = 3 atm, P2 = 7 atm, Cp = 28.8 J k-1 mol-1, T1 = 400 K, T2 = ?, Cp = 28.8 J k-1 mol-1, n = 2.07

Find T2 using P1 1-v T1 v = P2 1-v T2 v where v = 1.4

T2 will be 509.56

For adiabatic process Del (S) = 0

For Del Q (heat) = 2.07 x 28.8 x (509.56 - 400) = 6531.52 J

Del S (heat) = nCpln(T1/T2) = 14.5 JK-1

Del Q (comprsn) = nRT ln(P1/P2) = -7430.39 J

Del S (comprsn) = -7430.39/509.56 = -14.5 JK-1

So total entropy change = 0

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