(D) The number of moles in the first vessel is $\mu_1 = \frac{P_1V}{RT_1}$.The number of moles in the second vessel is $\mu_2 = \frac{P_2V}{RT_2}$.Whe

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$\frac{P_1}{2T_1} + \frac{P_2}{2T_2}$

(D) The number of moles in the first vessel is $\mu_1 = \frac{P_1V}{RT_1}$.The number of moles in the second vessel is $\mu_2 = \frac{P_2V}{RT_2}$.When connected,the total number of moles $\mu$ remains conserved,so $\mu = \mu_1 + \mu_2$.The total volume becomes $2V$,and the final pressure and temperature are $P$ and $T$. Thus,$\mu = \frac{P(2V)}{RT}$.Equating the moles: $\frac{P(2V)}{RT} = \frac{P_1V}{RT_1} + \frac{P_2V}{RT_2}$.Dividing both sides by $V/R$,we get: $\frac{2P}{T} = \frac{P_1}{T_1} + \frac{P_2}{T_2}$.Therefore,$\frac{P}{T} = \frac{P_1}{2T_1} + \frac{P_2}{2T_2}$.

Class 11 Physics Kinetic Theory of Gases Mixture of Gases

Difficult

Two vessels of equal volume contain the same gas. Their pressures and temperatures are $P_1, T_1$ and $P_2, T_2$ respectively. When they are connected,the common pressure and temperature are $P$ and $T$. Then $P/T$ is equal to:

A
$\frac{P_1}{T_1} + \frac{P_2}{T_2}$
B
$\frac{P_1T_1 + P_2T_2}{(T_1 + T_2)^2}$
C
$\frac{P_1T_2 + P_2T_1}{(T_1 + T_2)^2}$
D
$\frac{P_1}{2T_1} + \frac{P_2}{2T_2}$

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Kinetic Theory of Gases

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Mixture of Gases

In a container of volume $16.62 \ m^3$ at $0 \ ^{\circ}C$ temperature,$2 \ moles$ of oxygen,$5 \ moles$ of nitrogen,and $3 \ moles$ of hydrogen are present. Then,the pressure in the container is (Universal gas constant $R = 8.31 \ J \ mol^{-1} \ K^{-1}$) (in $Pa$)

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If a mixture of $28 \, g$ of Nitrogen $(N_2)$,$4 \, g$ of Hydrogen $(H_2)$,and $8 \, g$ of Helium $(He)$ is contained in a vessel at temperature $400 \, K$ and pressure $8.3 \times 10^5 \, Pa$,the density of the mixture will be:

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$A$ vessel of volume $0.02 \, m^3$ contains a mixture of hydrogen and helium at $20 \, ^oC$ and $2$ atmospheric pressure. The mass of the mixture is $5 \, g$. Find the ratio of the mass of hydrogen to that of helium in the mixture.

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$A$ vessel of volume $20 \ L$ contains a mixture of hydrogen and helium at a temperature of $27^{\circ} C$ and pressure $2 \ atm$. The mass of the mixture is $5 \ g$. Assuming the gases to be ideal,the ratio of the mass of hydrogen to that of helium in the given mixture will be:

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$A$ mixture of hydrogen and oxygen has a volume of $2000 \; cm^{3}$,a temperature of $300 \; K$,a pressure of $100 \; kPa$,and a mass of $0.76 \; g$. The ratio of the number of moles of hydrogen to the number of moles of oxygen in the mixture is:

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Two vessels of equal volume contain the same gas. Their pressures and temperatures are $P_1, T_1$ and $P_2, T_2$ respectively. When they are connected,the common pressure and temperature are $P$ and $T$. Then $P/T$ is equal to:

Similar Questions

In a container of volume $16.62 \ m^3$ at $0 \ ^{\circ}C$ temperature,$2 \ moles$ of oxygen,$5 \ moles$ of nitrogen,and $3 \ moles$ of hydrogen are present. Then,the pressure in the container is (Universal gas constant $R = 8.31 \ J \ mol^{-1} \ K^{-1}$) (in $Pa$)

If a mixture of $28 \, g$ of Nitrogen $(N_2)$,$4 \, g$ of Hydrogen $(H_2)$,and $8 \, g$ of Helium $(He)$ is contained in a vessel at temperature $400 \, K$ and pressure $8.3 \times 10^5 \, Pa$,the density of the mixture will be:

$A$ vessel of volume $0.02 \, m^3$ contains a mixture of hydrogen and helium at $20 \, ^oC$ and $2$ atmospheric pressure. The mass of the mixture is $5 \, g$. Find the ratio of the mass of hydrogen to that of helium in the mixture.

$A$ vessel of volume $20 \ L$ contains a mixture of hydrogen and helium at a temperature of $27^{\circ} C$ and pressure $2 \ atm$. The mass of the mixture is $5 \ g$. Assuming the gases to be ideal,the ratio of the mass of hydrogen to that of helium in the given mixture will be:

$A$ mixture of hydrogen and oxygen has a volume of $2000 \; cm^{3}$,a temperature of $300 \; K$,a pressure of $100 \; kPa$,and a mass of $0.76 \; g$. The ratio of the number of moles of hydrogen to the number of moles of oxygen in the mixture is:

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