Question

Consider the cyclic process ABCA, shown in figure, performed on a sample of 2.0 mol of an ideal gas. A total of 1200 J of heat is withdrawn from the sample in the process. Find the work done by the gas during the part BC.

Answer 1

Given:-

Number of moles of the gas, n = 2

∆Q = − 1200 J    (Negative sign shows that heat is extracted out from the system)

∆U = 0 ..............(During cyclic process)

Using the first law of thermodynamics, we get

∆Q = ∆U + ∆W

⇒ −1200 = 0 + (W_AB + W_BC + W_CA)

Since the change in volume of the system applies on line CA, work done during CA will be zero.

From the ideal gas equation, we get

PV = nRT

P∆V = nR∆T

W = P∆V = nR∆T

⇒ ∆Q = ∆U + ∆W

⇒ −1200 = nR∆T + W_BC + 0

⇒ −1200 = 2 × 8.3 × 200 + W_BC

W_BC = − 400 × 8.3 − 1200

= − 4520 J