(A) The ideal gas equation is given by $PV = \mu RT$,where $\mu = \frac{m}{M}$ is the number of moles.Substituting $\mu$,we get $PV = \frac{m}{M}RT$.S

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

$\frac{P_1}{T_1 d_1} = \frac{P_2}{T_2 d_2}$

(A) The ideal gas equation is given by $PV = \mu RT$,where $\mu = \frac{m}{M}$ is the number of moles.Substituting $\mu$,we get $PV = \frac{m}{M}RT$.Since density $d = \frac{m}{V}$,we can write $V = \frac{m}{d}$.Substituting $V$ in the equation: $P(\frac{m}{d}) = \frac{m}{M}RT$.Simplifying,we get $P = \frac{d}{M}RT$,which implies $\frac{P}{dT} = \frac{R}{M} = \text{constant}$.Therefore,for two different states,we have $\frac{P_1}{d_1 T_1} = \frac{P_2}{d_2 T_2}$.

Class 11 Physics Kinetic Theory of Gases Gas Laws (Charles, Boyle's, Avagadro's, Gay Lussacs and Dalton's law) and Ideal gas Equation

Medium

The equation of an ideal gas in terms of pressure $(P)$,absolute temperature $(T)$,and density $(d)$ is:

A
$\frac{P_1}{T_1 d_1} = \frac{P_2}{T_2 d_2}$
B
$\frac{P_1 T_1}{d_1} = \frac{P_2 T_2}{d_2}$
C
$\frac{P_1 d_2}{T_1} = \frac{P_2 d_1}{T_2}$
D
$\frac{P_1}{d_1 T_1} = \frac{P_2}{d_2 T_2}$

Explore More

Browse All

Question Bank

All Subjects

Class 11 Questions

Class 11

Physics Questions

Chapter

Kinetic Theory of Gases

Subtopic

Gas Laws (Charles, Boyle's, Avagadro's, Gay Lussacs and Dalton's law) and Ideal gas Equation

$A$ jar $P$ is filled with a gas having pressure,volume,and temperature $P, V, T$ respectively. Another gas jar $Q$ is filled with a gas having pressure $2P$,volume $\frac{V}{4}$,and temperature $2T$. The ratio of the number of molecules in jar $P$ to those in jar $Q$ is:

DifficultMHT CET 2023

View Solution

The gas which obeys Boyle's law for the maximum range of temperature is

Easy

View Solution

When the temperature of a gas in a closed vessel is increased by $1 \ ^\circ C$,its pressure increases by $0.4 \ \%$. The initial temperature of the gas is:

Difficult

View Solution

At constant pressure,the ratio of the increase in volume of an ideal gas per degree rise in Kelvin temperature to its original volume is $(T =$ absolute temperature of the gas$)$

Medium

View Solution

$A$ flask containing air at $27\,^{\circ}C$ at atmospheric pressure is corked up. $A$ pressure of $2.5\,atm$ would force the cork out. The temperature at which it happens is

Difficult

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 equation of an ideal gas in terms of pressure $(P)$,absolute temperature $(T)$,and density $(d)$ is:

Similar Questions

$A$ jar $P$ is filled with a gas having pressure,volume,and temperature $P, V, T$ respectively. Another gas jar $Q$ is filled with a gas having pressure $2P$,volume $\frac{V}{4}$,and temperature $2T$. The ratio of the number of molecules in jar $P$ to those in jar $Q$ is:

The gas which obeys Boyle's law for the maximum range of temperature is

When the temperature of a gas in a closed vessel is increased by $1 \ ^\circ C$,its pressure increases by $0.4 \ \%$. The initial temperature of the gas is:

At constant pressure,the ratio of the increase in volume of an ideal gas per degree rise in Kelvin temperature to its original volume is $(T =$ absolute temperature of the gas$)$

$A$ flask containing air at $27\,^{\circ}C$ at atmospheric pressure is corked up. $A$ pressure of $2.5\,atm$ would force the cork out. The temperature at which it happens is

Vedclass Test Series

Exam Paper Generator

Online Exam Module