In an adiabatic process,the initial and final temperatures of $1 \, mol$ of an ideal gas are $T_1$ and $T_2$ respectively. The change in internal energy of the gas is .......
- A$\frac{R}{\gamma - 1}(T_2 - T_1)$
- B$\frac{R}{\gamma - 1}(T_1 - T_2)$
- C$R(T_1 - T_2)$
- DZero
Explore More
Similar Questions
The figure shows an adiabatic curve on a $\log T$ and $\log V$ scale for an ideal gas. The gas is ............
Medium
View SolutionThe phenomenon of sound propagation in air is
Easy
View Solution$A$ diatomic gas undergoes an adiabatic change. Its pressure $P$ and temperature $T$ are related as $P \propto T^{x}$, where $x$ is:
$A$ given system undergoes a change in which the work done by the system equals the decrease in its internal energy. The system must have undergone an
$A$ gas $(\gamma = 1.3)$ is enclosed in an insulated vessel fitted with an insulating piston at a pressure of $10^5 \ N/m^2$. On suddenly pressing the piston,the volume is reduced to half the initial volume. The final pressure of the gas is:
Medium
View SolutionVedclass Products
For Students
Vedclass Test Series
Mock tests in real JEE/NEET style with performance analysis. 5-day free trial.
Start Free TrialFor Teachers
Exam Paper Generator
Generate Set A/B/C/D exam papers from 7.5L+ questions in 2 minutes. 3 chapters free.
Try FreeFor Institutes
Online Exam Module
Live online exams with unlimited students, 360° analytics & white-label branding.
See Demo