Question
Question: One mole of ideal monatomic gas (\(\gamma =\)5/3) is mixed with one mole of diatomic gas (\(\gamma=\...
One mole of ideal monatomic gas (γ=5/3) is mixed with one mole of diatomic gas (γ=7/5). γ denotes the ratio of specific heat at constant pressure, to that at constant volume. Find γ for the mixture.
A. 3/2
B. 23/15
C. 35/23
D. 4/3
Solution
The degrees of freedom for a monatomic gas is 3 and that for a diatomic gas is 5.
The ratio of specific heats γ for an ideal gas is
γ=1+f2
Let N1 moles of an ideal gas with f1 degrees of freedom per molecule be mixed with N2 moles of another ideal gas with f2 degrees of freedom per molecule at a particular temperature. The ratio of specific heats γ for the mixture is
γ=N1f1+N2f2N1(2+f1)+N2(2+f2)
Complete step by step answer:
For the monatomic gas,
Number of moles, N1=1
γ=35
Calculate the degrees of freedom f1 of the monatomic gas by substituting the value of γ in the γ-f relation:
\dfrac{5}{3}=1+\dfrac{2}{{{f}_{1}}} \\\
⟹35−1=f12
⟹32=f12
⟹f1=3
For the diatomic gas,
Number of moles, N2=1
γ=57
Calculate the degrees of freedom f2 of the monatomic gas by substituting the value of γ in the γ-f relation:
\dfrac{7}{5}=1+\dfrac{2}{{{f}_{2}}} \\\
⟹57−1=f22
⟹52=f22
⟹f2=5
Now, substituting the values of N1 ,N2and f1 ,f2 in the formula for the ratio of specific heats γ for the mixture:
γ=(1)(3)+(1)(5)1(2+31)+1(2+5)
⟹γ=812
⟹γ=23
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.