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Question: Calculate the specific heat capacity \[{C_v}\] of a gaseous mixture consisting of \[{v_1}\] moles of...

Calculate the specific heat capacity Cv{C_v} of a gaseous mixture consisting of v1{v_1} moles of gas of adiabatic exponent γ1{\gamma _1} and v2{v_2} moles of another gas of adiabatic exponent γ2{\gamma _2}.
(A) Cv=Rγ−1=Rv1+v2(v1γ1+1+v2γ2−1){C_v} = \dfrac{R}{{\gamma - 1}} = \dfrac{R}{{{v_1} + {v_2}}}\left( {\dfrac{{{v_1}}}{{{\gamma _1} + 1}} + \dfrac{{{v_2}}}{{{\gamma _2} - 1}}} \right)
(B) Cv=Rγ−1=Rv1−v2(v1γ1−1+v2γ2−1){C_v} = \dfrac{R}{{\gamma - 1}} = \dfrac{R}{{{v_1} - {v_2}}}\left( {\dfrac{{{v_1}}}{{{\gamma _1} - 1}} + \dfrac{{{v_2}}}{{{\gamma _2} - 1}}} \right)
(C) Cv=Rγ−1=Rv1+v2(v1γ1+1+v2γ2−1){C_v} = \dfrac{R}{{\gamma - 1}} = \dfrac{R}{{{v_1} + {v_2}}}\left( {\dfrac{{{v_1}}}{{{\gamma _1} + 1}} + \dfrac{{{v_2}}}{{{\gamma _2} - 1}}} \right)
(D) Cv=Rγ−1=Rv1+v2(v1γ1−1+v2γ2−1){C_v} = \dfrac{R}{{\gamma - 1}} = \dfrac{R}{{{v_1} + {v_2}}}\left( {\dfrac{{{v_1}}}{{{\gamma _1} - 1}} + \dfrac{{{v_2}}}{{{\gamma _2} - 1}}} \right)

Explanation

Solution

The specific heat capacity Cv{C_v} for any gas sample is equal to Rγ−1\dfrac{R}{{\gamma - 1}} where R is the universal gas constant and γ\gamma is the adiabatic exponent of the gas. Another thing to notice and remember is that the gases are adiabatic, which means that upon mixing, their internal energies would not change. Let’s see how it’s done.

Complete step-by-step solution:
We know that, Cv=Rγ−1{C_v} = \dfrac{R}{{\gamma - 1}}
And internal energy of a gas, U=PVγ−1U = \dfrac{{PV}}{{\gamma - 1}}
Let the volume of each gas taken be V litres.
Now, U1=P1Vγ1−1{U_1} = \dfrac{{{P_1}V}}{{{\gamma _1} - 1}} and U2=P2Vγ2−1{U_2} = \dfrac{{{P_2}V}}{{{\gamma _2} - 1}}
When adiabatic gases are mixed, the internal energies of the gases are conserved (as discussed in the hint). This is the characteristic property of adiabatic gases.
The internal energy of the gas mixture, U=U1+U2U = {U_1} + {U_2}
Hence, PVγ−1=P1Vγ1−1+P2Vγ2−1\dfrac{{PV}}{{\gamma - 1}} = \dfrac{{{P_1}V}}{{{\gamma _1} - 1}} + \dfrac{{{P_2}V}}{{{\gamma _2} - 1}}
Taking volume out of both sides of the equation, we get Pγ−1=P1γ1−1+P2γ2−1\dfrac{P}{{\gamma - 1}} = \dfrac{{{P_1}}}{{{\gamma _1} - 1}} + \dfrac{{{P_2}}}{{{\gamma _2} - 1}}-----equation 1
Ideal gas equations for the mixture of gases are as follows:
PV=(v1+v2)RTPV = ({v_1} + {v_2})RT
Now, ideal gas equations for the unmixed gases will be:
P1V=v1RT{P_1}V = {v_1}RT
P2V=v2RT{P_2}V = {v_2}RT
From the above-given equations, we can say that P1=v1v1+v2P{P_1} = \dfrac{{{v_1}}}{{{v_1} + {v_2}}}P and P2=v2v1+v2P{P_2} = \dfrac{{{v_2}}}{{{v_1} + {v_2}}}P
Substituting the above values in equation 1, we get 1γ−1=1v1+v2(v1γ1−1+v2γ2−1)\dfrac{1}{{\gamma - 1}} = \dfrac{1}{{{v_1} + {v_2}}}\left( {\dfrac{{{v_1}}}{{{\gamma _1} - 1}} + \dfrac{{{v_2}}}{{{\gamma _2} - 1}}} \right)
Cv=Rγ−1=Rv1+v2(v1γ1−1+v2γ2−1){C_v} = \dfrac{R}{{\gamma - 1}} = \dfrac{R}{{{v_1} + {v_2}}}\left( {\dfrac{{{v_1}}}{{{\gamma _1} - 1}} + \dfrac{{{v_2}}}{{{\gamma _2} - 1}}} \right)

Additional Information: For an adiabatic process, PVγP{V^\gamma }= constant and for an ideal gas Cp−Cv=R{C_p} - {C_v} = R . These two equations can help solve a lot of questions for you. Also, keep in mind that CpCv=γ\dfrac{{{C_p}}}{{{C_v}}} = \gamma and you won’t have to worry.

Note:- For individual gas samples, this becomes very easy as they have their adiabatic exponents but when we are given a mixture of gases, we need to find the adiabatic exponent of the gas mixture first.