Question

Two mole of a perfect gas undergo the following processes: 

(a) a reversible isobaric expansion from (1.0 atm, 20.0L) to (1.0 atm, 40.0L). 

(b) a reversible isochoric change of state from (1.0 atm, 40.0L) to (0.5 atm, 40.0L) 

(c) a reversible isothermal compression from (0.5 atm, 40.0L) to (1.0 atm, 20.0L). 

(i) sketch with labels each of the processes on the same P-V diagram. 

(ii) Calculate the total work (w) and the total heat change (q) involved in the above processes. 

(iii) What will be the value of U, H and S for the overall process ?

Answer 1

The overall is cyclic one, i.e., initial state is regained, thus

ΔU = 0; ΔH = 0 and ΔS = 0. 

Now, total work W = W_{A→ B} + W_{B→ C} + W_{C→ A}

W_AB = – P (V_B – V_A )

= – 1 (40 – 20) = – 20L atm

= – 20 × 1.01325 × 10² J

= – 2026.5 J

W_BC = O (Isochoric)

W_CA = – 2.303 nRT log₁₀ V_A/V_C

n = 2 mol.

At point C : P = 0.5 atm, V = 40L

PV = nRT

T = (0.5 x 40)/(0.0821) (2) = 121.8 K

W_CA = – 2.303 (2) (8.314) (121.8) log₁₀ (20/40)

= 1404.07 J

Total work, W = – 2026.5 + 0 + 1404.07

= – 622.43 J

For cyclic process : ΔU = 0

⇒ q = – w

q = + 622.43 J