Q70 / 79 · NEST (NISER) 2026 · Physics
One mole of a monatomic ideal gas undergoes a transformation from an initial state with temperature $290$ K and volume $30$ litres to a final state with temperature $310$ K and volume $16$ litres. On the pressure–volume ($P - V$) diagram, this process is represented by a straight line path. The magnitude of the work done (in joules) during this process is close to
- A$1939$
- B$877$
- C$1375$
- Correct answerD$1690$
Solution
For a straight line path on the $P-V$ diagram, the work done is given by the area of the trapezium under the curve:
$W = \dfrac{1}{2}(P_1 + P_2)(V_2 - V_1)$
The magnitude of the work done is:
$|W| = \dfrac{1}{2}(P_1 + P_2)|V_2 - V_1|$
Using the ideal gas equation $P = \dfrac{nRT}{V}$ for $n = 1$ mole:
$P_1 = \dfrac{RT_1}{V_1} = \dfrac{8.314 \times 290}{30 \times 10^{-3}} = 80.368 \times 10^3 \text{ Pa}$
$P_2 = \dfrac{RT_2}{V_2} = \dfrac{8.314 \times 310}{16 \times 10^{-3}} = 161.083 \times 10^3 \text{ Pa}$
The change in volume is:
$|V_2 - V_1| = |16 - 30| \times 10^{-3} = 14 \times 10^{-3} \text{ m}^3$
Substituting these values into the work done formula:
$|W| = \dfrac{1}{2} (80.368 \times 10^3 + 161.083 \times 10^3) \times 14 \times 10^{-3}$
$|W| = \dfrac{1}{2} (241.451 \times 10^3) \times 14 \times 10^{-3}$
$|W| = 241.451 \times 7 \approx 1690 \text{ J}$
Answer: $1690$
$W = \dfrac{1}{2}(P_1 + P_2)(V_2 - V_1)$
The magnitude of the work done is:
$|W| = \dfrac{1}{2}(P_1 + P_2)|V_2 - V_1|$
Using the ideal gas equation $P = \dfrac{nRT}{V}$ for $n = 1$ mole:
$P_1 = \dfrac{RT_1}{V_1} = \dfrac{8.314 \times 290}{30 \times 10^{-3}} = 80.368 \times 10^3 \text{ Pa}$
$P_2 = \dfrac{RT_2}{V_2} = \dfrac{8.314 \times 310}{16 \times 10^{-3}} = 161.083 \times 10^3 \text{ Pa}$
The change in volume is:
$|V_2 - V_1| = |16 - 30| \times 10^{-3} = 14 \times 10^{-3} \text{ m}^3$
Substituting these values into the work done formula:
$|W| = \dfrac{1}{2} (80.368 \times 10^3 + 161.083 \times 10^3) \times 14 \times 10^{-3}$
$|W| = \dfrac{1}{2} (241.451 \times 10^3) \times 14 \times 10^{-3}$
$|W| = 241.451 \times 7 \approx 1690 \text{ J}$
Answer: $1690$
Hence, the correct answer is option: 'D'.