Physics · 2019
JEE Main · 11 January 2019, Shift 1 · Q10
A rigid diatomic ideal gas undergoes an adiabatic process at room temperature. The relation between temperature and volume for this process is TV^x=…
A rigid diatomic ideal gas undergoes an adiabatic process at room temperature. The relation between temperature and volume for this process is $\displaystyle \mathrm{TV}^x=$ constant, then $\displaystyle x$ is :
Official answer
(2)
$\displaystyle \frac{2}{5}$
More from Thermodynamics
- The volume of an ideal gas (γ=1.5) is changed adiabatically from 5 litres to 4 litres. The ratio of initial pressure to final pressure is:2024
- The pressure P_1 and density d_1 of diatomic gas (γ=7/5) changes suddenly to P_2(>P_1) and d_2 respectively during an adiabatic process. The…2022
- A certain amount of gas of volume V at 27°C temperature and pressure 2 × 10^7 Nm^-2 expands isothermally until its volume gets doubled. Later it…2022
- The magnitude of heat exchanged by a system for the given cyclic process ABCA (as shown in figure) is (in SI unit):2025
- The pressure of a gas changes linearly with volume from A to B as shown in figure. If no heat is supplied to or extracted from the gas then change in…2023
- If one mole of an ideal gas at (P_1, V_1) is allowed to expand reversibly and isothermally (A to B) its pressure is reduced to one-half of the…2021
- A heat engine operates between a cold reservoir at temperature T_2=400 K and a hot reservoir at temperature T_1. It takes 300 J of heat from the hot…2021
- A monoatomic gas at pressure P and volume V is suddenly compressed to one eighth of its original volume. The final pressure at constant entropy will…2022
JEE Main 2019 Physics question, with the answer from NTA’s final answer key. Where our answers come from.