(B) The ideal gas equation is given by $PV = nRT$,where $n = \frac{m}{M}$.Thus,$PV = \frac{m}{M}RT$,which implies $m = \frac{PVM}{RT}$.For container $

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(B) The ideal gas equation is given by $PV = nRT$,where $n = \frac{m}{M}$.Thus,$PV = \frac{m}{M}RT$,which implies $m = \frac{PVM}{RT}$.For container $A$:$m_A = \frac{PVM}{RT} \quad (1)$For container $B$:$m_B = \frac{(2P)(2V)M}{R(T/2)} = \frac{4PVM}{RT/2} = \frac{8PVM}{RT} \quad (2)$Taking the ratio of $m_A$ to $m_B$:$\frac{m_A}{m_B} = \frac{PVM/RT}{8PVM/RT} = \frac{1}{8}$.Therefore,the ratio is $1: 8$.

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

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$A$ container $A$ contains an ideal gas at pressure $P$,volume $V$,and temperature $T$. $A$ second container $B$ contains the same gas at pressure $2P$,volume $2V$,and temperature $\frac{T}{2}$. The ratio of the mass of gas in $A$ to that in $B$ is:

MHT CET 2022 All MHT CET

A
$1: 4$
B
$1: 8$
C
$1: 1$
D
$1: 2$

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Class 11

Physics Questions

Chapter

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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Two vessels separately contain two ideal gases $A$ and $B$ at the same temperature. The pressure of $A$ is twice that of $B.$ Under such conditions,the density of $A$ is found to be $1.5$ times the density of $B.$ The ratio of the molecular weight of $A$ to that of $B$ is

MediumAIPMT 2015

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$A$ pressure $P-$absolute temperature $T$ diagram was obtained when a given mass of gas was heated. During the heating process from state $1$ to state $2$,the volume:

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The temperature of a gas at $-173^{\circ} C$ is $1 \text{ atm}$ at constant volume. Up to what temperature should the gas be heated so that the pressure becomes $2 \text{ atm}$ (in $^{\circ} C$)?

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The following graph represents the $T-V$ curves of an ideal gas (where $T$ is the temperature and $V$ is the volume) at three pressures $P_1, P_2$ and $P_3$ compared with those of Charles's law represented as dotted lines. Then the correct relation is:

MediumNEET 2024

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If the intermolecular forces vanish away,the volume occupied by the molecules contained in $4.5 \,kg$ of water at standard temperature and pressure will be

Medium

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$A$ container $A$ contains an ideal gas at pressure $P$,volume $V$,and temperature $T$. $A$ second container $B$ contains the same gas at pressure $2P$,volume $2V$,and temperature $\frac{T}{2}$. The ratio of the mass of gas in $A$ to that in $B$ is:

Similar Questions

Two vessels separately contain two ideal gases $A$ and $B$ at the same temperature. The pressure of $A$ is twice that of $B.$ Under such conditions,the density of $A$ is found to be $1.5$ times the density of $B.$ The ratio of the molecular weight of $A$ to that of $B$ is

$A$ pressure $P-$absolute temperature $T$ diagram was obtained when a given mass of gas was heated. During the heating process from state $1$ to state $2$,the volume:

The temperature of a gas at $-173^{\circ} C$ is $1 \text{ atm}$ at constant volume. Up to what temperature should the gas be heated so that the pressure becomes $2 \text{ atm}$ (in $^{\circ} C$)?

The following graph represents the $T-V$ curves of an ideal gas (where $T$ is the temperature and $V$ is the volume) at three pressures $P_1, P_2$ and $P_3$ compared with those of Charles's law represented as dotted lines. Then the correct relation is:

If the intermolecular forces vanish away,the volume occupied by the molecules contained in $4.5 \,kg$ of water at standard temperature and pressure will be

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