A sample of an ideal gas is taken through the cyclic process ABCA as shown in figure. It absorbs, 40 J of heat during the part AB, no heat during BC and rejects 60 J of heat during CA. A work of 50 J is done on the gas during the part BC. The internal energy of the gas at A is 1560 J. The workdone by the gas during the part CA is :

Let's analyze the cyclic process ABCA for the ideal gas.

Given Data:

• Heat absorbed during process AB (Q_AB): 40 J
• Heat absorbed during process BC (Q_BC): 0 J
• Heat rejected during process CA (Q_CA): -60 J (negative because it's rejected)
• Work done on the gas during BC (W_BC): -50 J (work done on the gas is taken as negative)
• Internal energy at point A (U_A): 1560 J

Step 1: Calculate Net Heat (ΔQ) for the Cycle

Since it's a cyclic process, the net change in internal energy (ΔU) over one complete cycle is zero.

Using the first law of thermodynamics:

ΔQ_cycle = ΔW_cycle

Net heat (ΔQ) absorbed during the cycle:

ΔQ_cycle = Q_AB + Q_BC + Q_CA

Substitute values:

ΔQ_cycle = 40 J + 0 J - 60 J = -20 J

So, the net heat absorbed, ΔQ_cycle, is -20 J.

Step 2: Calculate Net Work Done (ΔW) for the Cycle

Since ΔU for the cycle is zero, we can equate:

ΔQ_cycle = ΔW_cycle

This implies:

ΔW_cycle = -20 J

Step 3: Determine Work Done in Process CA (W_CA)

We know that:

ΔW_cycle = W_BC + W_CA

Substitute values:

-20 J = -50 J + W_CA

Solve for W_CA:

W_CA = -20 J + 50 J = 30 J

Final Answer

The work done by the gas during the process CA is 30 J.