AIIMS2018Physics-Thermodynamics

AIIMS 2018 Physics Adiabatic Processes MCQ Question

Type: MCQ-conceptual-Medium-Class 11

One mole of an ideal gas at an initial temperature of T K does 6R joules of work adiabatically. If the ratio of specific heats of this gas at constant pressure and at constant volume is 5/3, the final temperature of gas will be

A

(T − 4) K

B

(T + 2.4) K

C

(T − 2.4) K

D

(T + 4) K

Correct Answer

Option A

Detailed Explanation

In an adiabatic process for an ideal gas, the relationship between temperature and work done can be expressed using the formula W=CVCPCV(TfTi)W = \frac{C_V}{C_P - C_V} (T_f - T_i), where CPC_P and CVC_V are the specific heats at constant pressure and volume, respectively. Given that the ratio of specific heats CPCV=53\frac{C_P}{C_V} = \frac{5}{3}, we find CV=3R2C_V = \frac{3R}{2} and CP=5R2C_P = \frac{5R}{2}. The work done, W=6RW = 6R, leads to a final temperature Tf=T4T_f = T - 4 K, confirming option A as correct. Other options are incorrect as they do not satisfy the derived relationship for the final temperature based on the work done in this adiabatic process.

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