Two ideal monoatomic and diatomic gases are mixed are mixed at same temperature with one another to form an ideal gas mixture- The equation of the adiabatic process of the mixture is P V Y - constant- where Y -11-7 - If n 1 and n 2 are the number of moles of the mono atomic and diatomic gases in the mixture respectively- find the ratio n 1- n 2 Answer upto two digit after decimal point-

As, γ is given, we can find f using following relation: γ=1+2/f γ = 11/7 = 1 + 4/7⇒ f=7/2 The calculated f is for the mixture of gases, so f=f_mixture We know ...

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Standard XII

Chemistry

Irreversible Adiabatic Expansion And Compression

Two ideal mon...

Question

Two ideal monoatomic and diatomic gases are mixed are mixed at same temperature with one another to form an ideal gas mixture. The equation of the adiabatic process of the mixture is $P V^{γ} =$ constant, where $γ = \frac{11}{7} .$ If $n_{1}$ and $n_{2}$ are the number of moles of the mono atomic and diatomic gases in the mixture respectively, find the ratio $\frac{n_{1}}{n_{2}}$ (Answer upto two digit after decimal point).

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Solution

As, $γ$ is given, we can find $f$ using following relation:
$γ = 1 + \frac{2}{f}$
$γ = \frac{11}{7} = 1 + \frac{4}{7} \Rightarrow f = \frac{7}{2}$
The calculated $f$ is for the mixture of gases, so $f = f_{m i x t u r e}$

We know that, if two different types of gases are mixed at same temperature together, then the degree of freedom of the mixture is calculated by:
$f_{m i x t u r e} = \frac{n_{1} . f_{1} + n_{2} . f_{2}}{n_{1} + n_{2}}$
Here,
$f_{1} =$ Degree of freedom of gas $1$ =3
$f_{2} =$ Degree of freedom of gas $2$ =5

$f_{m i x t u r e} = \frac{7}{2} = \frac{n_{1} . 3 + n_{2} .5}{n_{1} + n_{2}} \Rightarrow \frac{n_{1}}{n_{2}} = 3$

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