JEE Main · 2025 · 24 Jan 2025 · Shift 1 · Heat & Thermodynamics

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 ______

Answer: 15

Solution

For a constant-pressure process on

n = 1

mole of an ideal monoatomic gas with temperature change

Δ T = 5 0^{\circ}

C:

Heat added:

E_{1} = Q = n C_{p} Δ T

Change in internal energy:

E_{2} = Δ U = n C_{v} Δ T

\frac{E _{1}}{E _{2}} = \frac{C _{p}}{C _{v}} = γ

For a monoatomic gas, degrees of freedom

f = 3

, so

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

Given

\frac{E _{1}}{E _{2}} = \frac{x}{9}

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

JEE Main · 2025 · 24 Jan 2025 · Shift 1 · Integer answer

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