A sample of 1.00 mol of an ideal gas (c_v, m = 5/2 R) at a pressure of 0.500 bar

Question

A sample of 1.00 mol of an ideal gas (c_v, m = 5/2 R) at a pressure of 0.500 bar and 250. K is compressed adiabatically by a constant external pressure of 1.00 bar until the internal pressure is equal to 1.00 bar. Determine the final temperature, and volume. For the same gas, determine the final pressure temperature and volume if the gas were reversibly and adiabatically compressed to a final pressure of 1.00 bar. What reversible process is required to add to b to leave the gas in the same state as in Part a? Calculate the heat transferred and the entropy change for the process you chose. Calculate the change in entropy for the process in Part a. Keep in mind that this is an irreversible process. The Claussius inequality states that d_qirr lessthanorequalto TdS or that dq_irr lessthanorequalto dq_rev. Show that this is true for this problem by comparing q in Part a to q in Parts b and c.

Explanation / Answer

a) Adiabatic irreversible compression

T2 = [Cv +R P2/P1]T1

Cp-Cv = R

Cp = 5R/2

5R/2 - Cv = R

3R/2 = Cv

T2 =( [3R/2 +Rx1/0.500]250)/5R/2 = 350 K

Initial volume can be calculated using ideal gas equation

PV = nRT ; V = nRT/P = 1.000 mol x .08314 L bar mol-1 K-1 x 250 K /0.500 bar

V1= 41.57 L

V2 = 1.000 mol x .08314 L bar mol-1 K-1 x 350 K /1.000 bar = 29.099 L

b) T2 = T1(P2/P1)(-1)/   Where = Cp/Cv = (5R/2)/(3R/2) = 5/3

(-1)/ = (5/3 - 1) / 5/3 =

T2 = 250 K(1.000/.500)2/5 = 250 x (2)0.4 = 250 x 1.320 = 329.87 K = 330 K

To calculate volume V2

V2 = V1(P1/P2)1/ = 41.57 L (0.500/ 1.000)3/5 = 41.57*0.6598 = 27.4 L

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