Question

Two moles of ideal monoatomic gas go through a cyclic transformation DABCD staring at the point D in the diagram below at a pressure of 2 atmospheres and a temperature of 360K and returning at point D. The volume at B is three times that of point D. The gas pressure at B is twice the gas pressure at C. Paths AB and CD represent isothermal processes.

a) Calculate the net work done by the gas during one cycle.

Answer 1

As given in the question,

Pressure at point D: P_D = 2 atm = 2*101325 Pa = 202650 Pa

Temperature at point D: T_D = 360 K,   no of moles: n = 2 mol

Volume ratio: V_B = 3*V_D   and Pressure ratio: P_B = 2*P_C

Using ideal gas equation, P_D*V_D = n*R*T_D

=>   V_D  = n*R*T_D / P_D

    = 2*8.314*360 / 202650 = 0.0295 m^3

V_A = V_D =  0.0295 m^3

So, the volume at point B: V_B = 3*V_D  = 0.0886 m^3

   V_C = V_B =  0.0886 m^3

Now, the work done by gas can be calculated using below formula:

Isothermal process: W = - n*R*T*ln(V_final / V_initial)

Temperatures, T_C = T_D = 360 K

Ideal gas equation between points B and C,

n*R*T_B / P_B*V_B = n*R*T_C / P_C*V_C

Since, V_C  = V_B and P_B = 2*P_C => T_B  = 2*T_C  = 2*360 = 720 K

  T_A = T_B = 720 K

The work done in (D -> A) and (B -> C) are zero because these are at constant volume.

So, the net work done in the whole cyclic process,

W_Total  = W(A -> B) + W(C -> D)

   = {- n*R*T_A*ln(V_B / V_A)} + {- n*R*T_C*ln(V_D / V_C)}

= {- 2*8.314*720*ln(0.0886 / 0.0295)} + {- 2*8.314*360*ln(0.0295 / 0.0886)}

= - 13166.28 J + 6583.14 J = - 6583.14 J = - 6.583*10^3 kJ