Question:

Given below are two statements: one is labelled as Assertion A and the other is labelled as Reason R}
Statement I: Change in internal energy of a system containing $n$ mole of ideal gas can be written as $\Delta U = n C_v (T_f - T_i) = \frac{nR}{\gamma - 1}(T_f - T_i)$, where $\gamma = \frac{C_p}{C_v}, T_i = \text{initial temperature}, T_f = \text{final temperature}$.
Statement II: Relation between degree of freedom $f$ and $\gamma (= C_p/C_v)$ is $\gamma = 1 + \frac{2}{f}$}
Choose the correct answer from the options given below

Updated On: Apr 12, 2026
  • Both A and R are true and R is the correct explanation of A
  • Both A and R are true but R is NOT the correct explanation of A
  • A is true but R is false
  • A is false but R is true
Hide Solution
collegedunia
Verified By Collegedunia

The Correct Option is A

Solution and Explanation

Step 1: Understanding the Question:
We need to verify the validity of the internal energy formula for an ideal gas and the relationship between the ratio of specific heats ($\gamma$) and degrees of freedom ($f$).
Step 2: Detailed Explanation:
For an ideal gas, the change in internal energy is given by:
\[ \Delta U = n C_v \Delta T \]
From Mayer's relation $C_p - C_v = R$ and $\gamma = C_p/C_v$, we can write:
\[ C_v = \frac{R}{\gamma - 1} \]
Substituting this into the $\Delta U$ equation:
\[ \Delta U = \frac{nR}{\gamma - 1}(T_f - T_i) \]
Thus, Statement I is true.
Now, for degrees of freedom $f$, the molar specific heat at constant volume is $C_v = \frac{f}{2}R$.
Using $\gamma = \frac{C_p}{C_v} = \frac{C_v + R}{C_v} = 1 + \frac{R}{C_v}$:
\[ \gamma = 1 + \frac{R}{\frac{f}{2}R} = 1 + \frac{2}{f} \]
Thus, Statement II is true.
Since the expression for $C_v$ in Statement I depends directly on the relationship derived in Statement II, Statement II is the correct theoretical explanation for Statement I.
Step 3: Final Answer:
Both are true and Statement II explains Statement I.
Was this answer helpful?
0
0

Top JEE Main The Kinetic Theory of Gases Questions

View More Questions

Top JEE Main Questions

View More Questions