A rigid diatomic ideal gas undergoes an... | Thermodynamics

JEE MAIN 2019

11-01-2019 S2

Question

A rigid diatomic ideal gas undergoes an adiabatic process at room temperature. The relation between temperature and volume for this process is TV $\mathrm{x}=$ constant, then x is

Select the correct option:

$\frac{2}{5}$

$\frac{2}{3}$

$\frac{5}{3}$

$\frac{3}{5}$

✓ Correct! Well done.

✗ Incorrect. Try again or view the solution.

Solution

Question Tags

JEE Main

Physics

Medium

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