The volume of an ideal gas $(\gamma=1.5)$ is changed adiabatically from $5 \ L$ to $4 \ L$. The ratio of initial pressure to final pressure is:

  • A
    $4/5$
  • B
    $16/25$
  • C
    $8/(5\sqrt{5})$
  • D
    $2/\sqrt{5}$

Explore More

Similar Questions

Five moles of Hydrogen gas initially at $STP$ is compressed adiabatically so that its temperature becomes $673 \, K$. The increase in internal energy of the gas is $(R=8.3 \, J \, mol^{-1} \, K^{-1}, \gamma=1.4$ for diatomic gas$)$ (in $kJ$)

$1 \, \text{kmol}$ of an ideal gas is compressed adiabatically,requiring $146 \, \text{kJ}$ of work. During this process,the temperature of the gas increases by $7^o \text{C}$. The gas is: $(R = 8.3 \, \text{J mol}^{-1} \text{K}^{-1})$

At $27^\circ C$,a gas is suddenly compressed such that its pressure becomes $1/8$th of the original pressure. The temperature of the gas will be $(\gamma = 5/3)$.

In an adiabatic change,the pressure $P$ and temperature $T$ of a monoatomic gas are related by the relation $P \propto T^C$,where $C$ equals

$A$ gas consisting of rigid diatomic molecules was initially under standard conditions $(T_1 = 273.15 \, K)$. Then,the gas was compressed adiabatically to one-fifth of its initial volume. What will be the mean kinetic energy of a rotating molecule in the final state?

Vedclass Products

For Students

Vedclass Test Series

Mock tests in real JEE/NEET style with performance analysis. 5-day free trial.

Start Free Trial
For Teachers

Exam Paper Generator

Generate Set A/B/C/D exam papers from 7.5L+ questions in 2 minutes. 3 chapters free.

Try Free
For Institutes

Online Exam Module

Live online exams with unlimited students, 360° analytics & white-label branding.

See Demo