QJEEAdv18Phb6f9: JEE Advanced 2018 Physics - Class 11 JEEAdv18Phb6f9 Solution: One mole of a monatomic ideal gas undergoes four thermodynamic…
| LIST - I | LIST - II | ||
|---|---|---|---|
| P. | In process I | 1. | Work done by the gas is zero |
| Q. | In process II | 2. | Temperature of the gas remains unchanged |
| R. | In process III | 3. | No heat is exchanged between the gas and its surroundings |
| S. | In process IV | 4. | Work done by the gas is 6P0V0 |
Multiple Choice Options
- A
- B
- C
- D
Step-by-Step Verified Solution
Process I is adiabatic, therefore, Q = 0
Q = U + W W = U
Thus, W < 0 since volume of gas is decreasing U > 0 Therefore, temperature of gas increases and no heat is exchanged between gas and surroundings.
Thus, the correct mapping is P 3.
Process II is isobaric, therefore, pressure remains constant
W = P V = 3P0[3V0 V0] = 6P0V0
Therefore, work done by gas in process II is 6P0V0.
Thus, the correct mapping is Q 4.
Process III is isochoric, therefore, volume remains constant
Q = U + W Q = U
and work done by gas is zero.
Thus, the correct mapping is R 1.
Process IV is isothermal process, therefore temperature remains constant
Q = U + W Q = W and = 0
Thus, the correct mapping is S 2.
Therefore, option (C) is correct.