Question

A sample of an ideal gas is taken through the cyclic process ABCA as shown in figure below. It absorbs \(60\,J\) of heat during the part AB and rejects \(80\,J\) of heat during CA. There is no heat exchanged during the process BC. A work of \(40\,J\) is done on the gas during the part BC. If the internal energy of the gas at A is \(1450\,J\), then the work done by the gas during the part CA is:

• A. \(10\,J\)
• B. \(20\,J\)
• C. \(40\,J\)
• D. \(30\,J\)

Hint:

For cyclic thermodynamic processes, apply \(Q=\Delta U+W\) carefully for each path and remember that work done on gas is negative of work done by gas.

Answer 1

The Correct Option is B

Step 1: Use first law of thermodynamics.
\[ Q = \Delta U + W \]
Here, \(W\) is work done by the gas.

Step 2: Analyze process AB.
From the \(P-V\) diagram, AB is vertical, so volume remains constant.
\[ W_{AB}=0 \]
Given heat absorbed during AB is:
\[ Q_{AB}=60\,J \]
Therefore,
\[ \Delta U_{AB}=60\,J \]

Step 3: Find internal energy at B.
\[ U_A=1450\,J \]
\[ U_B=U_A+\Delta U_{AB} \]
\[ U_B=1450+60=1510\,J \]

Step 4: Analyze process BC.
Given no heat is exchanged during BC:
\[ Q_{BC}=0 \]
Work of \(40\,J\) is done on the gas, so work done by the gas is:
\[ W_{BC}=-40\,J \]
Using first law:
\[ 0=\Delta U_{BC}+(-40) \]
\[ \Delta U_{BC}=40\,J \]

Step 5: Find internal energy at C.
\[ U_C=U_B+\Delta U_{BC} \]
\[ U_C=1510+40=1550\,J \]

Step 6: Analyze process CA.
Given heat is rejected during CA:
\[ Q_{CA}=-80\,J \]
Internal energy change during CA is:
\[ \Delta U_{CA}=U_A-U_C \]
\[ \Delta U_{CA}=1450-1550=-100\,J \]

Step 7: Calculate work done by gas during CA.
Using first law:
\[ Q_{CA}=\Delta U_{CA}+W_{CA} \]
\[ -80=-100+W_{CA} \]
\[ W_{CA}=20\,J \]
\[ \boxed{20\,J} \] Hence, correct answer is option (B).