Physics · Thermodynamics

JEE Main 2025 — 23 January, Morning Shift — Question 65

item An ideal gas initially at $0^{\circ} \mathrm{C}$ temperature, is compressed suddenly to one fourth of its volume. If the ratio of specific heat at constant pressure to that at constant volume is $3 / 2$ , the change in temperature due to the thermodynamics process is $\qquad$ K.

Answer: 273

Numerical answer — enter this value.

Step-by-step solution

$\gamma=\frac{3}{2}$

$\mathrm{T}^{\gamma-1}=\mathrm{C}$

$273 \mathrm{~V}_{0}^{0.5}=\mathrm{T}\left(\frac{\mathrm{V}_{0}}{4}\right)^{0.5}$

$\mathrm{T}=273 \times 2=546$

$\Delta \mathrm{T}=273$

Answer key and solution verified before publishing.

Practise Thermodynamics

Start with this question, then two more from the same chapter — with a tutor that explains every step. Free.

Solve a similar one free

Exam: JEE Main 2025
Paper: 23 January, Morning Shift
Subject: Physics
Chapter: Thermodynamics
Topic: Different Thermodynamic Processes

Match the List-I with List-II List -I List -II --- --- --- --- A Pressure varies inversely with volume of an ideal gas. I Adiabatic process…

Answer: 273

Step-by-step solution

Practise Thermodynamics

More Thermodynamics questions from this paper