JEE Main 2025 Jan 24 Shift 1, Physics Q25: Thermodynamic Processes

JEE Main2025Jan 24, Shift 1Physics

The temperature of 1 mole of an ideal monoatomic gas is increased by $5 0^{\circ} C$ 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 _______.

Your answer

Solution

For an isobaric process: $E_{1} = Q = n C_{P} Δ T$ and $E_{2} = Δ U = n C_{V} Δ T$ . The ratio is

$\frac{E _{1}}{E _{2}} = \frac{C _{P}}{C _{V}} = γ$

For a monoatomic ideal gas, $f = 3$ degrees of freedom and

$γ = 1 + \frac{2}{f} = 1 + \frac{2}{3} = \frac{5}{3}$

Setting $\frac{x}{9} = \frac{5}{3}$ gives $x = \frac{5 \cdot 9}{3} = 15$ .

$∴$ Answer: $15$ .

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JEE Main 2025 Jan 24 Shift 1, Physics Q25: Thermodynamic Processes

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