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JEE Advance - Physics (2018 - Paper 2 Offline - No. 16)

One mole of a monatomic ideal gas undergoes four thermodynamic processes as shown schematically in the $$PV$$-diagram below. Among these four processes, one is isobaric, one is isochoric, one is isothermal and one is adiabatic. Match the processes mentioned in List-I with the corresponding statements in List-II.

JEE Advanced 2018 Paper 2 Offline Physics - Heat and Thermodynamics Question 49 English

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
$$P \to 4;Q \to 3;R \to 1;S \to 2$$
$$P \to 1;Q \to 3;R \to 2;S \to 4$$
$$P \to 3;Q \to 4;R \to 1;S \to 2$$
$$P \to 3;Q \to 4;R \to 2;S \to $$

Explanation

Process I is adiabatic, therefore, $$\Delta$$Q = 0

$$\Rightarrow$$ $$\Delta$$Q = $$\Delta$$U + W $$\Rightarrow$$ W = $$-$$$$\Delta$$U

Thus, W < 0 since volume of gas is decreasing $$\Rightarrow$$ $$\Delta$$U > 0 Therefore, temperature of gas increases and no heat is exchanged between gas and surroundings.

Thus, the correct mapping is P $$\to$$ 3.

Process II is isobaric, therefore, pressure remains constant

$$\Rightarrow$$ W = P$$\Delta$$V = 3P0[3V0 $$-$$ V0] = 6P0V0

Therefore, work done by gas in process II is 6P0V0.

Thus, the correct mapping is Q $$\to$$ 4.

Process III is isochoric, therefore, volume remains constant

$$\Rightarrow$$ $$\Delta$$Q = $$\Delta$$U + W $$\Rightarrow$$ $$\Delta$$Q = $$\Delta$$U

and work done by gas is zero.

Thus, the correct mapping is R $$\to$$ 1.

Process IV is isothermal process, therefore temperature remains constant

$$\Rightarrow$$ $$\Delta$$Q = $$\Delta$$U + W $$\Rightarrow$$ $$\Delta$$Q = W and $$\Delta$$ = 0

Thus, the correct mapping is S $$\to$$ 2.

Therefore, option (C) is correct.

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