An ideal gas is allowed to expand both reversibly and irreversibly in an isolated system. If T_i is the initial temperature and T_f is the final tempe - Chemistry MCQ AIEEE 2006 2026 | ExamTest.live

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An ideal gas is allowed to expand both reversibly and irreversibly in an isolated system. If T_{i} is the initial temperature and T_{f} is the final temperature, which of the following statements is correct?

A $(T_{f})_{i rre v} > (T_{f})_{re v}$
B $T_{f} > T_{i} for reversible process but T_{f} = T_{i} for irreversible process$
C $(T_{f})_{re v} = (T_{f})_{i rre v}$
D $T_{f} = T_{i} for both reversible and irreversible processes$

Solution & Step-by-step Explanation

In an isolated system, q = 0 . For an expansion, work w is done by the system, so w is negative.From the First Law: Δ U = q + w = w .Since Δ U = n C_{v} Δ T, and w is negative, Δ T must be negative (temperature drops).In a reversible adiabatic expansion, the system does the maximum possible work (∣ w_{re v} ∣ > ∣ w_{i rre v} ∣).Therefore, the drop in internal energy and temperature is greater for the reversible process. (T_{i} - T_{f})_{re v} > (T_{i} - T_{f})_{i rre v} (T_{f})_{i rre v} > (T_{f})_{re v}

Try it yourself before checking the explanation above.

An ideal gas is allowed to expand both reversibly and irreversibly in an isolated system. If T_{i} is the initial temperature and T_{f} is the final temperature, which of the following statements is correct?

(T_{f})_{i rre v} > (T_{f})_{re v}

T_{f} > T_{i} for reversible process but T_{f} = T_{i} for irreversible process

(T_{f})_{re v} = (T_{f})_{i rre v}

T_{f} = T_{i} for both reversible and irreversible processes

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