2.15 mol of an ideal gas with CV,m=3R/2 undergoes the transformations described in the following list from an initial state described by T=350.K and P=5.00bar.
1) The gas undergoes a reversible adiabatic expansion until the final pressure is one-fourth its initial value.
2) The gas undergoes an adiabatic expansion against a constant external pressure of 1.25 bar until the final pressure is one-fourth its initial value.
3)The gas undergoes an expansion against a constant external pressure of zero bar until the final pressure is equal to one-fourth of its initial value.
a) Calculate q for each process. (q1,q2,q3)
b) Calculate w for each process. (w1,w2,w3)
c) Calculate ΔU for each process. (ΔU1, ΔU2, ΔU3)
d) Calculate ΔH for each process. (ΔH1, ΔH2, ΔH3)
e) Calculate ΔS for each process. (ΔS1, ΔS2, ΔS3)
a) In the adiabatic process, there will be no heat exchange between the system and surroundings.
i.e. q1 = 0
b) w1 = n * CV,m * (T2 - T1)
T2/T1 = (P2/P1)(-1)/
Here, = 5/3 (since CV,m = 3/2 R)
Therefore, T2 = 350 K * (1/4)2/5 = 201 K
Hence, w1 = 2.15 mol * 3/2 * 8.3145 J mol-1 K-1 * (201-350) K = -3.995 kJ
c) For an adiabatic reversible process, w = U
Therefore, U1 = -3.995 kJ
d) H = 5/3 U (since H = 5/2 R and U = 3/2 R)
Therefore, H1 = 5/3 * -3.995 kJ = -6.658 kJ
e) S1 = 0 (since q1 = 0, reversible process)
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