Q157 / 180 · COMEDK 2024 (Afternoon Shift) · Physics
A thermodynamic system changes its state from A to C in two different paths ABC and AC. The internal energy of the system at state C is $20\text{ J}$ and at state A is $10\text{ J}$. Heat supplied to the system to go from A to C along path AC is
- Azero
- B60 J
- C90 J
- Correct answerD70 J
Solution
$\Delta U = U_C - U_A = 20 - 10 = 10$ J (state function, path independent)
Work done along path AC (area of trapezium under P-V curve):
$W_{AC} = \dfrac{1}{2}(P_A + P_C)(V_C - V_A) = \dfrac{1}{2}(5+15)(9-3) = \dfrac{1}{2} \times 20 \times 6 = 60$ J
By first law: $Q_{AC} = \Delta U + W_{AC} = 10 + 60 = 70$ J
Work done along path AC (area of trapezium under P-V curve):
$W_{AC} = \dfrac{1}{2}(P_A + P_C)(V_C - V_A) = \dfrac{1}{2}(5+15)(9-3) = \dfrac{1}{2} \times 20 \times 6 = 60$ J
By first law: $Q_{AC} = \Delta U + W_{AC} = 10 + 60 = 70$ J
Hence, the correct answer is option: 'D'.