Class JEE Chemistry ALL Q #1188
COMPETENCY BASED
APPLY
4 Marks 2026 JEE Main 2026 (Online) 21st January Morning Shift MCQ SINGLE
Find the change in internal energy of a gas if its temperature changes by 10K. Moles=10, $C_{p}=7 cal K^{-1}mol^{-1}$, $R=2~cal~K^{-1}mol^{-1}$.
(A) $500 cal$
(B) $1000 cal$
(C) $250 cal$
(D) $100 cal$
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AI Tutor Explanation

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Step-by-Step Solution

**Step 1: Find $C_v$** We know that $C_p - C_v = R$. Therefore, $C_v = C_p - R = 7 - 2 = 5~cal~K^{-1}mol^{-1}$.
**Step 2: Calculate the change in internal energy ($\Delta U$)** The change in internal energy is given by $\Delta U = n C_v \Delta T$, where $n$ is the number of moles and $\Delta T$ is the change in temperature. $\Delta U = 10 \times 5 \times 10 = 500~cal$.

Correct Answer: $500 cal$

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AI Suggestion: Option A

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Pedagogical Audit
Bloom's Analysis: This is an APPLY question because the student needs to apply the formula relating $C_p$, $C_v$, and the change in temperature to calculate the change in internal energy.
Knowledge Dimension: PROCEDURAL
Justification: The question requires the student to follow a specific procedure or algorithm to arrive at the solution, involving the application of a formula and the correct substitution of values.
Syllabus Audit: In the context of JEE, this is classified as COMPETENCY. The question assesses the student's ability to apply thermodynamic principles to solve a numerical problem, which goes beyond simple recall of definitions and formulas.

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