In an adiabatic expansion of a gas, the initial and final temperatures are $T_1$ and $T_2$ respectively. Then, the change in internal energy of the gas is:
$[R = \text{gas constant}, \gamma = \text{adiabatic ratio}]$

  • A
    zero
  • B
    $\frac{nR}{\gamma-1}(T_1-T_2)$
  • C
    $\frac{nR}{\gamma-1}(T_2-T_1)$
  • D
    $nR(T_1-T_2)$

Explore More

Similar Questions

In an adiabatic process,if the pressure is increased by $\frac{2}{3}\% $ and the ratio of specific heats $\frac{C_p}{C_v} = \gamma = \frac{3}{2}$,then the volume decreases by about:

In an adiabatic expansion of a gas,

During an adiabatic process,the pressure of a gas is found to be proportional to the cube of its temperature. The ratio $C_P / C_V$ for this gas is ........

Difficult
View Solution

An ideal gas undergoes an adiabatic process from state $(P_1, V_1, T_1)$ to $(P_2, V_2, T_2)$. The work done during the process is ..... (where $\mu$ = number of moles,$C_P$ and $C_V$ = molar specific heats).

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

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