Question

Three moles of an ideal gas perform a cycle as shown in the figure below. If T_A = 600 K, T_B = 800 K, T_C = 2,200 K and T_D = 1,200 K, calculate the work done along the cycle ABCD. Given R = 8.31 J mole^-1 K^-1.

Answer 1

Let p and p' be the pressure corresponding to the paths BC and DA.

The W_BC = p x (V_C - V_B)

= pV_C - pV_B

Now, pV_B = 3RT_B = 3R x 800

= 2,400 R (∵ μ = 3)

and pV_C = 3RT_C

= 3R x 2,200

= 6600 R

∴ W_BC = 6600 R - 2400 R = 4200 R

Again, W_DA = p'(V_A - V_D)

= p'V_A - p'V_D = 3 RT_A - 3RT_D

= 3R(T_A - T_D) = 3R (600 - 1200)

= -1800 R

The thermodynamic processes AB and CD are represented by straight lines. Hence, for both processes, p ∝ T i.e., Volume is constant.

Thus, there is no change in the volume of the gas along the path AB or CD.

W_AB = W_CD = 0

Hence, work done over the complete cycle,

W = W_AB + W_BC + W_CD + W_DA

= 0 + 4200 R + (-1800 R) + 0

= 2400 R = 2400 x 8.31

= 19,944 J