(A) According to Dalton's law of partial pressure, the total pressure $p$ of a mixture of gases is the sum of the partial pressures of the individual

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(A) According to Dalton's law of partial pressure, the total pressure $p$ of a mixture of gases is the sum of the partial pressures of the individual gases: $p = p_1 + p_2 + p_3$.Using the ideal gas equation $pV = nRT$, the pressure of each gas is given by $p_i = \frac{n_i RT}{V}$.Since the temperature $T$ and volume $V$ are the same for all gases in the mixture, the total pressure is:$p = \frac{n_1 RT}{V} + \frac{n_2 RT}{V} + \frac{n_3 RT}{V}$$p = \frac{(n_1 + n_2 + n_3) RT}{V}$.

Class 11 Physics Kinetic Theory Mixture of Gases

Easy

$A$ vessel of volume $V$ contains a mixture of ideal gases at temperature $T$. The gas mixture contains $n_1, n_2$ and $n_3$ moles of three gases. Assuming an ideal gas system, the pressure of the mixture is

TS EAMCET 2018 All TS EAMCET

A
$\frac{(n_1+n_2+n_3) R T}{V}$
B
$\frac{(n_1 n_2 n_3) R T}{V}$
C
$\frac{R T}{(n_1+n_2+n_3) V}$
D
$\frac{R T}{V(n_1 n_2 n_3)}$

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One mole of monoatomic gas and three moles of diatomic gas are put together in a container. The molar specific heat (in $J\,K^{-1}\,mol^{-1}$) at constant volume is $(R = 8.3\,J\,K^{-1}\,mol^{-1})$.

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The specific heat of the mixture of two gases at constant volume is $\frac{13}{6} R$. The ratio of the number of moles of the first gas to the second is $1:2$. The respective gases may be:

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$3\, \text{moles}$ of an ideal gas at a temperature of $27^{\circ}\, \text{C}$ are mixed with $2\, \text{moles}$ of an ideal gas at a temperature of $227^{\circ}\, \text{C}$. Determine the equilibrium temperature $(^{\circ}\, \text{C})$ of the mixture, assuming no loss of energy.

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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:

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Four moles of hydrogen,two moles of helium and one mole of water vapour form an ideal gas mixture. $[C_v$ for hydrogen $= \frac{5}{2} R, C_v$ for helium $= \frac{3}{2} R, C_v$ for water vapour $= 3 R]$. What is the molar specific heat at constant pressure of the mixture?

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$A$ vessel of volume $V$ contains a mixture of ideal gases at temperature $T$. The gas mixture contains $n_1, n_2$ and $n_3$ moles of three gases. Assuming an ideal gas system, the pressure of the mixture is

Similar Questions

One mole of monoatomic gas and three moles of diatomic gas are put together in a container. The molar specific heat (in $J\,K^{-1}\,mol^{-1}$) at constant volume is $(R = 8.3\,J\,K^{-1}\,mol^{-1})$.

The specific heat of the mixture of two gases at constant volume is $\frac{13}{6} R$. The ratio of the number of moles of the first gas to the second is $1:2$. The respective gases may be:

$3\, \text{moles}$ of an ideal gas at a temperature of $27^{\circ}\, \text{C}$ are mixed with $2\, \text{moles}$ of an ideal gas at a temperature of $227^{\circ}\, \text{C}$. Determine the equilibrium temperature $(^{\circ}\, \text{C})$ of the mixture, assuming no loss of energy.

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Four moles of hydrogen,two moles of helium and one mole of water vapour form an ideal gas mixture. $[C_v$ for hydrogen $= \frac{5}{2} R, C_v$ for helium $= \frac{3}{2} R, C_v$ for water vapour $= 3 R]$. What is the molar specific heat at constant pressure of the mixture?

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