Question

One mole of an ideal gas undergoes a cyclic process as shown in the figure.

Calculate the work done by the gas in each of the process. A-B, B-C, and C-A. What is the network done by the gas at the end of one cycle?

Answer 1

1. The process A - B is isobaric. Therefore work done is given by, 

W_AB = P. ∆V

= 200 × 10³ × (5-3) × 10^-3

= 400J 

∴ Also, work done is given by area under the curve, AB.

Area = (AB) × (200) × 10³

= (2 × 10^-3) × 200 × 10³ =400

∴ Area = W_AB.

2. Work done in the process BC,

W_BC = Area under the curve BC = 0

3. Work done in process CA,

W_CA= - (Area under the curve CA)

∴ W_CA = - [\(\frac{1}{2}\) × (100 + 200) × 10³ × (2 × 10^-3)]

∴ W_CA = - 300 J

∴ Total work done by the gas at the end of cycle,

W_total = (400 + 0 - 300) J = 100 J.