When 300 J of heat given to an ideal gas with C p =7/2 R its temperature raises from 20 ° C to 50 ° C keeping its volume constant. The mass of the gas is (approximately) \ \ \ \ g .( R =8.314 J / mol . K ) .

Physics · Thermodynamics

JEE Main 2026 — 24 January, Evening Shift — Question 44

When 300 J of heat given to an ideal gas with $\mathrm{C}_{\mathrm{p}}=\frac{7}{2} \mathrm{R}$ its temperature raises from $20^{\circ} \mathrm{C}$ to $50^{\circ} \mathrm{C}$ keeping its volume constant. The mass of the gas is (approximately) $\_\_\_\_$ $\mathrm{g} .(\mathrm{R}=8.314 \mathrm{~J} / \mathrm{mol} . \mathrm{K})$ .

Answer: 479

Numerical answer — enter this value.

Step-by-step solution

$\mathrm{C}_{\mathrm{v}}=\mathrm{C}_{\mathrm{P}}-\mathrm{R}=\frac{5}{2} \mathrm{R}$ $\Delta \mathrm{Q}=\mathrm{nC}_{\mathrm{V}} \Delta \mathrm{T}$ $300=\mathrm{n} \times \frac{5}{2} \times 8.314 \times 30$ $\mathrm{n}=0.48$ $\frac{\mathrm{m}}{\mathrm{M}}=0.48$ We cannot find mass (m) because molar mass (M) not given.

Answer key and solution verified before publishing.

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Exam: JEE Main 2026
Paper: 24 January, Evening Shift
Subject: Physics
Chapter: Thermodynamics
Topic: First Law of Thermodynamics