(C) For an adiabatic process, the relationship between pressure $P$ and density $d$ is given by $P \propto d^\gamma$.Given $\gamma = \frac{7}{5}$ and

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

(C) For an adiabatic process, the relationship between pressure $P$ and density $d$ is given by $P \propto d^\gamma$.Given $\gamma = \frac{7}{5}$ and $\frac{d^{\prime}}{d} = 32$.We have the relation $\frac{P^{\prime}}{P} = \left(\frac{d^{\prime}}{d}\right)^\gamma$.Substituting the given values:$\frac{P^{\prime}}{P} = (32)^{7/5}$.Since $32 = 2^5$, we have:$\frac{P^{\prime}}{P} = (2^5)^{7/5} = 2^7$.Calculating the value: $2^7 = 128$.Therefore, the ratio $\frac{P^{\prime}}{P}$ is $128$.

Class 11 Physics Thermodynamics Adiabatic Process

Difficult

The pressure and density of a diatomic gas $\left(\gamma=\frac{7}{5}\right)$ changes adiabatically from $(P, d)$ to $(P^{\prime}, d^{\prime})$. If $\frac{d^{\prime}}{d}=32$, then $\frac{P^{\prime}}{P}$ is:

TS EAMCET 2001 All TS EAMCET

AP EAMCET 2001 All AP EAMCET

A
$\frac{1}{128}$
B
$32$
C
$128$
D
$256$

Explore More

Browse All

Question Bank

All Subjects

Class 11 Questions

Class 11

Physics Questions

Chapter

Thermodynamics

Subtopic

Adiabatic Process

Previous Year

TS EAMCET 2001 Paper All TS EAMCET years →

Previous Year

AP EAMCET 2001 Paper All AP EAMCET years →

In the $P-V$ diagram shown,there are two adiabatic parts of the same gas intersecting two isothermals at $T_1$ and $T_2$. The ratio $\left(\frac{V_b}{V_a}\right)$ is equal to

DifficultMHT CET 2022

View Solution

$A$ gas having $\gamma = 5/2$ and volume $360 \text{ c.c.}$ is suddenly compressed to $90 \text{ c.c.}$ If the initial pressure of the gas is $P$,then the final pressure will be:

MediumMHT CET 2025

View Solution

Compressed air in the tube of a bicycle wheel at normal temperature suddenly starts coming out from a puncture. The air inside:

Easy

View Solution

$A$ monoatomic gas is stored in a thermally insulated container and the gas is suddenly compressed to $(\frac{1}{8})^{\text{th}}$ of its initial volume. The ratio of final pressure to initial pressure is (in $:1$)

DifficultMHT CET 2026

View Solution

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 $(\gamma = 5/3)$.

Medium

View Solution

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

The pressure and density of a diatomic gas $\left(\gamma=\frac{7}{5}\right)$ changes adiabatically from $(P, d)$ to $(P^{\prime}, d^{\prime})$. If $\frac{d^{\prime}}{d}=32$, then $\frac{P^{\prime}}{P}$ is:

Similar Questions

In the $P-V$ diagram shown,there are two adiabatic parts of the same gas intersecting two isothermals at $T_1$ and $T_2$. The ratio $\left(\frac{V_b}{V_a}\right)$ is equal to

$A$ gas having $\gamma = 5/2$ and volume $360 \text{ c.c.}$ is suddenly compressed to $90 \text{ c.c.}$ If the initial pressure of the gas is $P$,then the final pressure will be:

Compressed air in the tube of a bicycle wheel at normal temperature suddenly starts coming out from a puncture. The air inside:

$A$ monoatomic gas is stored in a thermally insulated container and the gas is suddenly compressed to $(\frac{1}{8})^{\text{th}}$ of its initial volume. The ratio of final pressure to initial pressure is (in $:1$)

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 $(\gamma = 5/3)$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module