JEE Advanced · 2014 · 25 May 2014 · Paper 1 · Chemical Thermodynamics

An ideal gas in a thermally insulated vessel at internal pressure

= P_{1}

, volume

= V_{1}

and absolute temperature

= T_{1}

expands irreversibly against zero external pressure, as shown in the diagram. The final internal pressure, volume and absolute temperature of the gas are

P_{2}, V_{2}

and

T_{2}

, respectively. For this expansion,

(A) $q = 0$ ✓ Correct
(B) $T_{2} = T_{1}$ ✓ Correct
(C) $P_{2} V_{2} = P_{1} V_{1}$ ✓ Correct
(D) $P_{2} V_{2}^{γ} = P_{1} V_{1}^{γ}$

Answer: (A), (B), (C)

Solution

This is a free expansion (expansion against zero external pressure) in a thermally insulated vessel (adiabatic walls).

• Since the vessel is thermally insulated, no heat is exchanged:

q = 0

(option A correct).
• Work done:

W = - P_{e x t} Δ V = 0

(since

P_{e x t} = 0

).
• By the first law:

Δ U = q + W = 0 + 0 = 0

.
• For an ideal gas,

Δ U = n C_{V} Δ T

. Since

Δ U = 0

, we have

Δ T = 0

, so

T_{2} = T_{1}

(option B correct).
• Since temperature is constant and the gas is ideal:

P_{1} V_{1} = n R T_{1}

and

P_{2} V_{2} = n R T_{2} = n R T_{1}

, so

P_{2} V_{2} = P_{1} V_{1}

(option C correct).
• Option (D)

P_{2} V_{2}^{γ} = P_{1} V_{1}^{γ}

applies only to reversible adiabatic processes. This expansion is irreversible, so (D) is incorrect.

Correct answers: (A), (B), (C).

JEE Advanced · 2014 · 25 May 2014 · Paper 1 · One or more correct options

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JEE Advanced · 2014 · 25 May 2014 · Paper 1 · Chemical Thermodynamics

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