(D) The ideal gas equation is given by $PV = NkT$,where $N$ is the number of molecules and $k$ is the Boltzmann constant.For jar $A$: $P_A = P$,$V_A =

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(D) The ideal gas equation is given by $PV = NkT$,where $N$ is the number of molecules and $k$ is the Boltzmann constant.For jar $A$: $P_A = P$,$V_A = V$,$T_A = T$. Thus,$N_A = \frac{PV}{kT}$.For jar $B$: $P_B = 2P$,$V_B = V/4$,$T_B = 2T$. Thus,$N_B = \frac{(2P)(V/4)}{k(2T)} = \frac{PV/2}{2kT} = \frac{PV}{4kT}$.The ratio of the number of molecules is $\frac{N_A}{N_B} = \frac{PV/kT}{PV/4kT} = \frac{4}{1}$.Therefore,the ratio is $4:1$.

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 pressure $P$,volume $V$,and temperature $T$ of a gas in jar $A$ are given. Another gas in jar $B$ has pressure $2P$,volume $V/4$,and temperature $2T$. The ratio of the number of molecules in jar $A$ to jar $B$ is:

AIIMS 1982 All AIIMS

A
$1:1$
B
$1:2$
C
$2:1$
D
$4:1$

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Class 11

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Kinetic Theory of Gases

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

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AIIMS 1982 Paper All AIIMS years →

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The pressure $P$,volume $V$,and temperature $T$ of a gas in jar $A$ are given. Another gas in jar $B$ has pressure $2P$,volume $V/4$,and temperature $2T$. The ratio of the number of molecules in jar $A$ to jar $B$ is:

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In a process,the pressure of an ideal gas varies with volume as $P = \frac{a}{1 + (V/b)^2}$,where $a$ and $b$ are constants. What is the temperature of the gas when the volume of $1$ mole of the gas is $V = b$?

$4$ moles of an ideal gas is at $0^\circ C$. At constant pressure,it is heated to double its volume. Its final temperature will be ...... $^\circ C$.

What is the equation of state for $8 \, g$ of $O_2$?

What is the mass of $2$ litres of nitrogen gas $(N_2)$ at $22.4$ atmospheric pressure and $273\,K$?

An ideal gas at atmospheric pressure has a temperature of $300 \,K$ and a volume of $1 \,m^3$. If its temperature and volume are both doubled,its pressure will be ...........

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