Question

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 work done by the gas during the part CA is

• A. 20J
• B. 30J
• C. -30J
• D. -60J

Answer 1

The Correct Option is B

To solve this problem, we need to apply the first law of thermodynamics, which states:

Q = \Delta U + W

where Q is the heat exchanged, \Delta U is the change in internal energy, and W is the work done by the system.

1. Heat and Work Information:
   • Heat absorbed during AB: Q_{AB} = 40 \, \text{J}
   • No heat exchange during BC: Q_{BC} = 0 \, \text{J}
   • Heat rejected during CA: Q_{CA} = -60 \, \text{J}
   • Work done on the gas during BC: W_{BC} = -50 \, \text{J}
2. Internal Energy Changes:
   • The initial internal energy at A is U_A = 1560 \, \text{J}.
3. Cyclic Nature:
   • Since the process is cyclic (ABCA), the total change in internal energy is zero: \Delta U_{cycle} = 0
4. Calculate Work Done During CA:

   For the entire cycle, using the first law:

   Q_{total} = W_{AB} + W_{BC} + W_{CA}

   Substituting the given heat exchanges:

   40 \, \text{J} + 0 \, \text{J} - 60 \, \text{J} = W_{AB} + (-50 \, \text{J}) + W_{CA}

   Simplifying gives:

   -20 \, \text{J} = W_{AB} - 50 \, \text{J} + W_{CA}

   Since W_{AB} is 0 in this cyclic analysis, resolving gives:

   W_{CA} = 30 \, \text{J}

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