QJEEM25Ph444a: JEE Main 2025 Physics - Class 11 JEEM25Ph444a Solution: The temperature of 1 mole of an ideal monoatomic gas is…

$Q = n\,C_P\,\Delta T \quad \text{and} \quad \Delta U = n\,C_V\,\Delta T$

For a monoatomic ideal gas, the molar specific heats are given by:

$C_V = \frac{3}{2}R \quad \text{and} \quad C_P = C_V + R = \frac{5}{2}R$

Thus, the ratio of the heat added to the change in internal energy is:

$\frac{Q}{\Delta U} = \frac{C_P}{C_V} = \frac{\frac{5}{2}R}{\frac{3}{2}R} = \frac{5}{3}$

Given that:

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

We equate the two ratios:

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

Multiplying both sides by 9:

$x = \frac{5}{3} \times 9 = 15$

Thus, the value of $x$ is 15.