$A$ 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:
- A$2^{0.7} \times 10^5 \ N/m^2$
- B$2^{1.3} \times 10^5 \ N/m^2$
- C$2^{1.4} \times 10^5 \ N/m^2$
- DNone of these
Explore More
Similar Questions
$A$ monoatomic ideal gas,initially at temperature $T_1$,is enclosed in a cylinder fitted with a frictionless piston. The gas is allowed to expand adiabatically to a temperature $T_2$ by releasing the piston suddenly. If $L_1$ and $L_2$ are the lengths of the gas column before and after expansion respectively,then $T_1/T_2$ is given by
DifficultIIT 2000
View SolutionThe figure shows an adiabatic curve on a $\log T$ and $\log V$ scale for an ideal gas. The gas is ............
Medium
View SolutionAn ideal gas at $27^{\circ}\text{C}$ is compressed adiabatically to $(8/27)$ of its original volume. If $\gamma = 5/3$, then the rise in temperature of a gas is (in $\text{ K}$)
DifficultMHT CET 2026
View SolutionThe initial pressure and volume of an ideal gas are $P_0$ and $V_0$. The final pressure of the gas when the gas is suddenly compressed to volume $\frac{V_0}{4}$ will be (Given $\gamma =$ ratio of specific heats at constant pressure and at constant volume).
$A$ reversible adiabatic path on a $P-V$ diagram for an ideal gas passes through state $A$ where $P = 0.7 \times 10^5 \, N/m^2$ and $V = 0.0049 \, m^3$. The ratio of specific heats of the gas is $1.4$. The slope of the path at $A$ is:
Difficult
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