Class NEET 2026 ALL Q #1916
COMPETENCY BASED
APPLY
4 Marks 2026 NTA-RE-NEET-2026 MCQ SINGLE
In an adiabatic expansion, the temperature of one mole of an ideal monatomic gas $(\gamma=5/3)$ decreases from 60 K to 50 K. The work done by the gas in the process is: (Take the universal gas constant as $R=8.3~J~mol^{-1}K^{-1})$
(A) 166 J
(B) 41.5 J
(C) 83 J
(D) 124.5 J
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Correct Answer: D

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Detailed Solution

Step 1: Identify the formula for work done in an adiabatic process

For an adiabatic process, the work done by an ideal gas is given by the change in internal energy with a negative sign, or specifically using the temperature change: $$W = \frac{nR(T_1 - T_2)}{\gamma - 1}$$

Step 2: List the given parameters

Given: $n = 1$ mole $T_1 = 60$ K $T_2 = 50$ K $\gamma = 5/3$ $R = 8.3$ J mol⁻¹ K⁻¹

Step 3: Calculate the denominator

The value of $(\gamma - 1)$ is: $$\gamma - 1 = \frac{5}{3} - 1 = \frac{2}{3}$$

Step 4: Substitute values into the work formula

Substitute the values into the equation: $$W = \frac{1 \times 8.3 \times (60 - 50)}{2/3}$$ $$W = \frac{8.3 \times 10}{2/3}$$ $$W = \frac{83 \times 3}{2}$$ $$W = \frac{249}{2} = 124.5 \text{ J}$$

Final Answer: 124.5 J

AI Suggestion: Option D

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Pedagogical Audit
Bloom's Analysis: This is an APPLY question because the student must select the appropriate thermodynamic formula for adiabatic work and perform numerical substitution.
Knowledge Dimension: PROCEDURAL
Justification: The question requires the application of a specific sequence of steps (formula selection, parameter identification, and arithmetic calculation) to solve a physics problem.
Syllabus Audit: In the context of NEET, this is classified as COMPETENCY. It tests the student's ability to apply thermodynamic principles to specific state changes, which is a core requirement for the NEET Physics syllabus.

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