Question

11. Consider a Carnot cycle with 2.25 moles of a diatomic ideal gas as the working substance (assume Cv = 2.5*R). The following are the steps of the cycle:

Step I: reversible, isothermal expansion at 300.0 °C from 10.00 L to 16.00 L.

Step II: reversible, a diabatic expansion until the temperature decreases to 50.0 °C.

Step III: reversible, isothermal compression at 50.0 °C.

Step IV: reversible, adiabatic compression back to the initial conditions.

A. Calculate q, w, ΔU, ΔH, and ΔS for each step and for the overall cycle; make a table of results.

B. Construct a computer-generated P-V plot of the cycle, labeling each step.

Answer 1

Part A.

Step I:

-w = nRT ln(V2/V1)

= 2.25 mol * 8.3145 J mol^-1 K^-1 * (300+273.15) K * ln(16 L/10 L)

= 5.04 KJ

i.e. w = -5.04 KJ

q = +5.04 KJ

i.e. TS = 5.04 KJ

i.e. S = 5.04 KJ/573.15 K = 8.79 J K^-1

Now, H = -w = +5.04 KJ

Step II:

w = U = {nR/(-1)} * T

= {2.25 mol * 8.3145 J mol^-1 K^-1 /(1.4-1)} * (50 - 300) K

i.e. w = -11.69 KJ

q = 0 KJ

i.e. TS = 0

i.e. S = 0 KJ

Now, q = U + (-w) = H = 0 KJ

Step III:

-w = nRT ln(V2/V1)

= 2.25 mol * 8.3145 J mol^-1 K^-1 * (50+273.15) K * ln(10 L/16 L)

= -2.84 KJ

i.e. w = +2.84 KJ

q = -2.84 KJ

i.e. TS = -2.84 KJ

i.e. S = -2.84 KJ/323.15 K = -8.79 J K^-1

Now, H = -w = -2.84 KJ

Step IV:

w = U = {nR/(-1)} * T

= {2.25 mol * 8.3145 J mol^-1 K^-1 /(1.4-1)} * (300 - 50) K

i.e. w = 11.69 KJ

q = 0 KJ

i.e. TS = 0

i.e. S = 0 KJ

Now, q = U + (-w) = H = 0 KJ

Part B.

I.E = isothermal expansion

A.E = Adiabatic expansion

I.C = Isothermal compression

A.C = Adiabatic compression

T1 = 300 ^oC, T2 = 50 ^oC

V1/V2 = V4/V3