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 A total of 1200 J of heat is withdrawn from the sample in the process.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.
Figure

Answer 1

The cyclic process ABCA, 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 A total of 1200 J of heat is withdrawn from the sample in the process. The work done by the gas during the part BC is -4520 J

Explanation:

Given that:  

n = 2

n is Amount of Gas Moles

∆Q = − 1200 J (-ve sign represent extracted heat from the system)

∆U = 0 (In cyclical processing)

Using the thermodynamics first law, we get

∆Q = ∆U + ∆W

⇒

Since the device changes in volume on line CA, Work done at CA is set to be zero.

The ideal gas equation, then we obtain

PV = nRT

P∆V = nR∆T

W = P∆V = nR∆T

∆Q = ∆U + ∆W

$-1200 = 2 \times200 \times 8.3+ W_{BC}$

$W_{BC}= - 400 \times 8.3 - 1200$

$W_{BC}= - 3320 - 1200$

       = - 4520 J