The temperature of 1 mole of an ideal monoatomic gas is increased… — Solution | JEE Physics

NVAEasyJEE 2025Degrees of Freedom & Law of Equipartition

The temperature of $1$ mole of an ideal monoatomic gas is increased by $50^\circ \text{C}$ at constant pressure. The total heat added and change in internal energy are $E_1$ and $E_2$ , respectively. If $\frac{E_1}{E_2} = \frac{x}{9}$ , then the value of $x$ is:

Answer

Correct answer:15

Step-by-step solution

Standard Method

Given: The process is isobaric and $\Delta T = 50^\circ \text{C}$ . The heat added is $E_1 = Q = nC_p\Delta T$ and the change in internal energy is $E_2 = \Delta U = nC_v\Delta T$ .

Find: The value of $x$ in $\frac{E_1}{E_2} = \frac{x}{9}$ .

For the ratio,

\frac{E_1}{E_2} = \frac{nC_p\Delta T}{nC_v\Delta T} = \frac{C_p}{C_v} = \gamma

For a monoatomic ideal gas,

\gamma = 1 + \frac{2}{f}

with $f = 3$ . Therefore,

\gamma = 1 + \frac{2}{3} = \frac{5}{3}

So,

\frac{E_1}{E_2} = \frac{5}{3}

Given that

\frac{E_1}{E_2} = \frac{x}{9}

we get

\frac{5}{3} = \frac{x}{9}

Hence,

x = 15

Therefore, the value of $x$ is $15$ .

Using heat capacities directly

Given: Isobaric heating of $1$ mole of a monoatomic ideal gas.

Find: The value of $x$ .

Use the standard values for a monoatomic gas:

C_p = \frac{5}{2}R, \qquad C_v = \frac{3}{2}R

Then,

\frac{E_1}{E_2} = \frac{nC_p\Delta T}{nC_v\Delta T} = \frac{C_p}{C_v} = \frac{\frac{5}{2}R}{\frac{3}{2}R} = \frac{5}{3}

Now compare with

\frac{x}{9} = \frac{5}{3}

So,

x = 15

This works because in the ratio, both $n$ and $\Delta T$ cancel out. Therefore, the value of $x$ is $15$ .

Common mistakes

Using $C_v$ for the heat added in an isobaric process is incorrect. At constant pressure, heat added is $Q = nC_p\Delta T$ . Use $C_p$ for heat and $C_v$ for change in internal energy.
Treating $50^\circ \text{C}$ differently from $50 \text{ K}$ for temperature change leads to confusion. For a temperature interval, $50^\circ \text{C} = 50 \text{ K}$ , although here the ratio makes $\Delta T$ cancel anyway.
Using the wrong value of $\gamma$ is a common conceptual error. For a monoatomic ideal gas, degrees of freedom are $f=3$ , so $\gamma = \frac{5}{3}$ , not $\frac{7}{5}$ which applies to a diatomic gas under usual conditions.

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