(A) The average kinetic energy per molecule of an ideal gas is given by the formula $K_{avg} = \frac{3}{2} k_{B} T$,where $k_{B}$ is the Boltzmann con

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Yes,they will have the same average kinetic energy per molecule.

(A) The average kinetic energy per molecule of an ideal gas is given by the formula $K_{avg} = \frac{3}{2} k_{B} T$,where $k_{B}$ is the Boltzmann constant and $T$ is the absolute temperature of the gas.Since both gases are in the same container,they are in thermal equilibrium and must be at the same temperature $T$.Because the average kinetic energy depends only on the temperature $T$ and not on the mass or nature of the gas molecules,both gases will have the same average kinetic energy per molecule.

Class 11 Physics Kinetic Theory Mixture of Gases

Medium

In a container,there is a mixture of two different gases. Will both gases have the same kinetic energy?

A
Yes,they will have the same average kinetic energy per molecule.
B
No,they will have different average kinetic energies.
C
Yes,they will have the same total kinetic energy.
D
No,the kinetic energy depends on the mass of the molecules.

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Two monoatomic ideal gases at temperatures $T_1$ and $T_2$ are mixed. There is no loss of energy. If the masses of molecules of the two gases are $m_1$ and $m_2$ and the number of their molecules are $n_1$ and $n_2$ respectively,the temperature of the mixture will be:

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$22 \ g$ of carbon dioxide at $27^{\circ} C$ is mixed in a closed container with $16 \ g$ of oxygen at $37^{\circ} C$. If both gases are considered as ideal gases,then the temperature of the mixture is nearly: (in $^{\circ} C$)

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Two ideal polyatomic gases at temperatures $T_{1}$ and $T_{2}$ are mixed such that there is no loss of energy. If $F_{1}$ and $F_{2}$,$m_{1}$ and $m_{2}$,$n_{1}$ and $n_{2}$ are the degrees of freedom,masses,and number of molecules of the first and second gas respectively,then the temperature of the mixture of these two gases is:

DifficultJEE MAIN 2021

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Two thermally insulated vessels $1$ and $2$ are filled with air at temperatures $T_1, T_2$,pressures $P_1, P_2$,and volumes $V_1, V_2$ respectively. If the valve connecting them is opened,what is the temperature of the equilibrium mixture?

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In a container,there is a mixture of two different gases. Will both gases have the same kinetic energy?

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Two monoatomic ideal gases at temperatures $T_1$ and $T_2$ are mixed. There is no loss of energy. If the masses of molecules of the two gases are $m_1$ and $m_2$ and the number of their molecules are $n_1$ and $n_2$ respectively,the temperature of the mixture will be:

The pressure of a mixture of $64 \ g$ of oxygen, $28 \ g$ of nitrogen, and $132 \ g$ of carbon dioxide gases in a closed vessel is $P$. Under isothermal conditions, if the entire oxygen is removed from the vessel, the pressure of the mixture of the remaining two gases is:

$22 \ g$ of carbon dioxide at $27^{\circ} C$ is mixed in a closed container with $16 \ g$ of oxygen at $37^{\circ} C$. If both gases are considered as ideal gases,then the temperature of the mixture is nearly: (in $^{\circ} C$)

Two thermally insulated vessels $1$ and $2$ are filled with air at temperatures $T_1, T_2$,pressures $P_1, P_2$,and volumes $V_1, V_2$ respectively. If the valve connecting them is opened,what is the temperature of the equilibrium mixture?

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