(C) The internal energy of the first gas is $U_1 = \frac{3}{2} n_1 R T_1$ (for monoatomic) or $U_1 = n_1 C_v T_1$.Since the problem implies a general

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$\frac{n_1 T_1 + n_2 T_2}{n_1 + n_2}$

(C) The internal energy of the first gas is $U_1 = \frac{3}{2} n_1 R T_1$ (for monoatomic) or $U_1 = n_1 C_v T_1$.Since the problem implies a general ideal gas mixture,the total internal energy is conserved.The total energy of the mixture is $U_{mix} = U_1 + U_2$.$U_{mix} = n_1 C_v T_1 + n_2 C_v T_2$.Let the final temperature be $T$. Then $U_{mix} = (n_1 + n_2) C_v T$.Equating the two: $(n_1 + n_2) C_v T = n_1 C_v T_1 + n_2 C_v T_2$.Canceling $C_v$ from both sides,we get $T = \frac{n_1 T_1 + n_2 T_2}{n_1 + n_2}$.

Class 11 Physics Kinetic Theory of Gases Mixture of Gases

Medium

Two ideal gases at temperatures $T_1$ and $T_2$ are mixed. There is no loss of energy. If the number of moles of the two gases are $n_1$ and $n_2$ respectively,then the temperature of the mixture will be:

A
$\frac{n_1 T_1 - n_2 T_2}{n_1 + n_2}$
B
$\frac{n_1 T_1 + n_2 T_2}{n_1 - n_2}$
C
$\frac{n_1 T_1 + n_2 T_2}{n_1 + n_2}$
D
$\frac{n_1 T_1 - n_2 T_2}{n_1 - n_2}$

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Kinetic Theory of Gases

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Mixture of Gases

$A$ gas mixture consists of $2.0$ moles of oxygen and $4.0$ moles of neon at temperature $T$. Neglecting all vibrational modes, calculate the total internal energy of the system. (Oxygen has two rotational modes.) (in $RT$)

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$A$ gaseous mixture contains an equal number of hydrogen $(H_2)$ and nitrogen $(N_2)$ molecules. Specific heat measurements on this mixture at temperatures below $100\, K$ would indicate that the value of $\gamma$ (ratio of specific heats) for this mixture is

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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:

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$1 \, g$ of $H_2$ at $27 \, ^oC$ is mixed with $16 \, g$ of $O_2$ at $37 \, ^oC$. The temperature of the mixture is about ....... $^oC$.

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$A$ vessel contains two non-reactive gases: neon (monatomic) and oxygen (diatomic). The ratio of their partial pressures is $3:2$. Estimate the ratio of
$(i)$ number of molecules and
$(ii)$ mass density of neon and oxygen in the vessel.
Atomic mass of $Ne = 20.2 \; u$,molecular mass of $O_2 = 32.0 \; u$.

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Two ideal gases at temperatures $T_1$ and $T_2$ are mixed. There is no loss of energy. If the number of moles of the two gases are $n_1$ and $n_2$ respectively,then the temperature of the mixture will be:

Similar Questions

$A$ gas mixture consists of $2.0$ moles of oxygen and $4.0$ moles of neon at temperature $T$. Neglecting all vibrational modes, calculate the total internal energy of the system. (Oxygen has two rotational modes.) (in $RT$)

$A$ gaseous mixture contains an equal number of hydrogen $(H_2)$ and nitrogen $(N_2)$ molecules. Specific heat measurements on this mixture at temperatures below $100\, K$ would indicate that the value of $\gamma$ (ratio of specific heats) for this mixture is

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:

$1 \, g$ of $H_2$ at $27 \, ^oC$ is mixed with $16 \, g$ of $O_2$ at $37 \, ^oC$. The temperature of the mixture is about ....... $^oC$.

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