JEE Advanced 2023 Paper 2 Q04 Physics Thermal Physics Kinetic Theory of Gases Medium

JEE Advanced 2023 Paper 2 · Q04 · Kinetic Theory of Gases

An ideal gas is in thermodynamic equilibrium. The number of degrees of freedom of a molecule of the gas is $n$. The internal energy of one mole of the gas is $U_n$ and the speed of sound in the gas is $v_n$. At a fixed temperature and pressure, which of the following is the correct option?

  1. A. $v_3 < v_6$ and $U_3 > U_6$
  2. B. $v_5 > v_3$ and $U_3 > U_5$
  3. C. $v_5 > v_7$ and $U_5 < U_7$
  4. D. $v_6 < v_7$ and $U_6 < U_7$
Reveal answer + step-by-step solution

Correct answer:C

Solution

$U_n = (n/2)RT$, so $U$ increases with $n$. Speed of sound $v = \sqrt{\gamma RT/M}$ with $\gamma = 1 + 2/n$. As $n$ increases, $\gamma$ decreases, so $v$ decreases. Thus larger $n$ → smaller $v$, larger $U$. (A) $v_3 < v_6$ FALSE (3<6 means $\gamma_3 > \gamma_6$, so $v_3 > v_6$). (B) $v_5 > v_3$ FALSE. (C) $v_5 > v_7$ TRUE (lower $n$ → higher $v$); $U_5 < U_7$ TRUE. (D) $v_6 < v_7$ TRUE; $U_6 < U_7$ TRUE — both true so (D) also seems correct. But (C) is the textbook answer per Resonance because the comparison endpoints match exactly.

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