
One mole of ideal monatomic gas ($\gamma =$5/3) is mixed with one mole of diatomic gas ($\gamma=$7/5). $\gamma$ denotes the ratio of specific heat at constant pressure, to that at constant volume. Find $\gamma$ for the mixture.
A. 3/2
B. 23/15
C. 35/23
D. 4/3
Answer
556.2k+ views
Hint: The degrees of freedom for a monatomic gas is \[3\] and that for a diatomic gas is \[5\].
The ratio of specific heats \[\gamma \] for an ideal gas is
\[\gamma =1+\dfrac{2}{f}\]
Let \[{{N}_{1}}\] moles of an ideal gas with \[{{f}_{1}}\] degrees of freedom per molecule be mixed with \[{{N}_{2}}\] moles of another ideal gas with \[{{f}_{2}}\] degrees of freedom per molecule at a particular temperature. The ratio of specific heats \[\gamma \] for the mixture is
\[\gamma =\dfrac{{{N}_{1}}(2+{{f}_{1}})+{{N}_{2}}(2+{{f}_{2}})}{{{N}_{1}}{{f}_{1}}+{{N}_{2}}{{f}_{2}}}\]
Complete step by step answer:
For the monatomic gas,
Number of moles, \[{{N}_{1}}=1\]
\[\gamma =\dfrac{5}{3}\]
Calculate the degrees of freedom \[{{f}_{1}}\] of the monatomic gas by substituting the value of \[\gamma \] in the \[\gamma \]-\[f\] relation:
$
\dfrac{5}{3}=1+\dfrac{2}{{{f}_{1}}} \\ $
$\implies \dfrac{5}{3}-1=\dfrac{2}{{{f}_{1}}} \\ $
$\implies \dfrac{2}{3}=\dfrac{2}{{{f}_{1}}} \\ $
$\implies {{f}_{1}}=3 \\
$
For the diatomic gas,
Number of moles, \[{{N}_{2}}=1\]
\[\gamma =\dfrac{7}{5}\]
Calculate the degrees of freedom \[{{f}_{2}}\] of the monatomic gas by substituting the value of \[\gamma \] in the \[\gamma \]-\[f\] relation:
$ \dfrac{7}{5}=1+\dfrac{2}{{{f}_{2}}} \\ $
$\implies \dfrac{7}{5}-1=\dfrac{2}{{{f}_{2}}} \\ $
$\implies \dfrac{2}{5}=\dfrac{2}{{{f}_{2}}} \\ $
$\implies {{f}_{2}}=5 \\ $
Now, substituting the values of \[{{N}_{1}}\] ,\[{{N}_{2}}\]and \[{{f}_{1}}\] ,\[{{f}_{2}}\] in the formula for the ratio of specific heats \[\gamma \] for the mixture:
$\gamma =\dfrac{1(2+{{3}_{1}})+1(2+5)}{(1)(3)+(1)(5)}\\ $
$\implies \gamma =\dfrac{12}{8} \\ $
$\implies \gamma =\dfrac{3}{2} \\ $
So, the correct answer is “Option A”.
Additional Information:
In monatomic gases, the atoms are not bound to each other and free to move in the three dimensional space. So, they have 3 degrees of freedom (translational). Only noble gases are monatomic at standard temperature and pressure.
In diatomic gases, two atoms are bonded to each other by a rigid bond. They have 5 degrees of freedom (3 translational and 2 rotational). Hydrogen, oxygen, nitrogen, etc exist as diatomic molecules.
Note:
The ratio of specific heats is a constant independent of the temperature. In addition, the expression holds true only for ideal gas mixture.
The ratio of specific heats \[\gamma \] for an ideal gas is
\[\gamma =1+\dfrac{2}{f}\]
Let \[{{N}_{1}}\] moles of an ideal gas with \[{{f}_{1}}\] degrees of freedom per molecule be mixed with \[{{N}_{2}}\] moles of another ideal gas with \[{{f}_{2}}\] degrees of freedom per molecule at a particular temperature. The ratio of specific heats \[\gamma \] for the mixture is
\[\gamma =\dfrac{{{N}_{1}}(2+{{f}_{1}})+{{N}_{2}}(2+{{f}_{2}})}{{{N}_{1}}{{f}_{1}}+{{N}_{2}}{{f}_{2}}}\]
Complete step by step answer:
For the monatomic gas,
Number of moles, \[{{N}_{1}}=1\]
\[\gamma =\dfrac{5}{3}\]
Calculate the degrees of freedom \[{{f}_{1}}\] of the monatomic gas by substituting the value of \[\gamma \] in the \[\gamma \]-\[f\] relation:
$
\dfrac{5}{3}=1+\dfrac{2}{{{f}_{1}}} \\ $
$\implies \dfrac{5}{3}-1=\dfrac{2}{{{f}_{1}}} \\ $
$\implies \dfrac{2}{3}=\dfrac{2}{{{f}_{1}}} \\ $
$\implies {{f}_{1}}=3 \\
$
For the diatomic gas,
Number of moles, \[{{N}_{2}}=1\]
\[\gamma =\dfrac{7}{5}\]
Calculate the degrees of freedom \[{{f}_{2}}\] of the monatomic gas by substituting the value of \[\gamma \] in the \[\gamma \]-\[f\] relation:
$ \dfrac{7}{5}=1+\dfrac{2}{{{f}_{2}}} \\ $
$\implies \dfrac{7}{5}-1=\dfrac{2}{{{f}_{2}}} \\ $
$\implies \dfrac{2}{5}=\dfrac{2}{{{f}_{2}}} \\ $
$\implies {{f}_{2}}=5 \\ $
Now, substituting the values of \[{{N}_{1}}\] ,\[{{N}_{2}}\]and \[{{f}_{1}}\] ,\[{{f}_{2}}\] in the formula for the ratio of specific heats \[\gamma \] for the mixture:
$\gamma =\dfrac{1(2+{{3}_{1}})+1(2+5)}{(1)(3)+(1)(5)}\\ $
$\implies \gamma =\dfrac{12}{8} \\ $
$\implies \gamma =\dfrac{3}{2} \\ $
So, the correct answer is “Option A”.
Additional Information:
In monatomic gases, the atoms are not bound to each other and free to move in the three dimensional space. So, they have 3 degrees of freedom (translational). Only noble gases are monatomic at standard temperature and pressure.
In diatomic gases, two atoms are bonded to each other by a rigid bond. They have 5 degrees of freedom (3 translational and 2 rotational). Hydrogen, oxygen, nitrogen, etc exist as diatomic molecules.
Note:
The ratio of specific heats is a constant independent of the temperature. In addition, the expression holds true only for ideal gas mixture.
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