JEE Advanced 2023 — Paper 2 — PHY-I solution

JEE Advanced2023Paper 2PHY-I

An ideal gas is in thermodynamic equilibrium. The number of degrees of freedom of a molecule of the gas in $n$ . 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?

A
$v_{3} < v_{6}$ and $U_{3} > U_{6}$
B
$v_{5} > v_{3}$ and $U_{3} > U_{5}$
C
$v_{5} > v_{7}$ and $U_{5} < U_{7}$
D
$v_{6} < v_{7}$ and $U_{6} < U_{7}$

Solution

For one mole of an ideal gas with $n$ degrees of freedom per molecule, the internal energy is $U_{n} = \frac{n}{2} R T$ . Since $T$ is fixed, $U_{n}$ increases monotonically with $n$ . This immediately rules out options that claim $U_{3} > U_{5}$ or $U_{3} > U_{6}$ .

The speed of sound in an ideal gas is

$v = \frac{γ R T}{M} = (1 + \frac{2}{n}) \frac{R T}{M}$ ,

so $v$ decreases as $n$ increases. Hence $v_{5} > v_{7}$ and $U_{5} < U_{7}$ , which is option (C).

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