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Question: Calculate the value of \( \gamma = \dfrac{{{C_p}}}{{{C_v}}} \) for a gaseous mixture containing of \...

Calculate the value of γ=CpCv\gamma = \dfrac{{{C_p}}}{{{C_v}}} for a gaseous mixture containing of v1=2.0{v_1} = 2.0 moles of
Oxygen and v2=3.0{v_2} = 3.0 moles of carbon-di-oxide. The gases are assumed to be ideal.
A. 1.33
B. 1.45
C. 2.11
D. 1.7

Explanation

Solution

Hint : Here we are asked to calculate the Heat capacity ratio of two gases i.e. oxygen and carbon dioxide. Cp{C_p} is heat capacity at constant pressure and Cv{C_v} is heat capacity at constant volume. Try to find the degree of freedom of these gases. Then calculate the given quantities of the given mixture of gases. Now you can easily calculate the value of γ\gamma .

Complete Step By Step Answer:
Let n1{n_1} be the moles of oxygen
n2{n_2} be the moles of carbon dioxide
Total moles = n1+n2=n{n_1} + {n_2} = n
γ1,Cp1,Cv1{\gamma _1},{C_{p1}},{C_{v1}} be for oxygen and
γ2,Cp2,Cv2{\gamma _2},{C_{p2}},{C_{v2}} be for carbon dioxide
Now we have to calculate the value of Cp{C_p} and Cv{C_v} for a given mixture of gases.
Cv{C_v} for mixture of gases.
Cv=Cv1n1+Cv2n2n1+n2{C_v} = \dfrac{{{C_{v1}}{n_1} + {C_{v2}}{n_2}}}{{{n_1} + {n_2}}} ---(1)
And Cp{C_p} for mixture of gases
Cp=Cp1n1+Cp2n2n1+n2{C_p} = \dfrac{{{C_{p1}}{n_1} + {C_{p2}}{n_2}}}{{{n_1} + {n_2}}} ----(2)
Now we have
γ=CpCv\gamma = \dfrac{{{C_p}}}{{{C_v}}} and Cv=Rγ−1{C_v} = \dfrac{R}{{\gamma - 1}}
Therefore Cp=γRγ−1{C_p} = \dfrac{{\gamma R}}{{\gamma - 1}}
Now putting all values in equation 1 and 2 we get
Cp=n1Cp1+n2Cp2n{C_p} = \dfrac{{{n_1}{C_{p1}} + {n_2}{C_{p2}}}}{n}
Cp=n1γ1Rγ1−1+n2γ2Rγ2−1n{C_p} = \dfrac{{{n_1}\dfrac{{{\gamma _1}R}}{{{\gamma _1} - 1}} + {n_2}\dfrac{{{\gamma _2}R}}{{{\gamma _2} - 1}}}}{n} ……… (3)
Cv=n1Cv1+n2Cv2n{C_v} = \dfrac{{{n_1}{C_{v1}} + {n_2}{C_{v2}}}}{n}
Cv=n1Rγ1−1+n2Rγ2−1n{C_v} = \dfrac{{{n_1}\dfrac{R}{{{\gamma _1} - 1}} + {n_2}\dfrac{R}{{{\gamma _2} - 1}}}}{n} ………. (4)
From equation 3 and 4 we get
γ=CpCv\gamma = \dfrac{{{C_p}}}{{{C_v}}} = =n1γ1Rγ1−1+n2γ2Rγ2−1nn1Rγ1−1+n2Rγ2−1n= \dfrac{{\dfrac{{{n_1}\dfrac{{{\gamma _1}R}}{{{\gamma _1} - 1}} + {n_2}\dfrac{{{\gamma _2}R}}{{{\gamma _2} - 1}}}}{n}}}{{\dfrac{{{n_1}\dfrac{R}{{{\gamma _1} - 1}} + {n_2}\dfrac{R}{{{\gamma _2} - 1}}}}{n}}}
Finally we get
γ=n1γ1(γ2−1)+n2γ2(γ1−1)n1(γ2−1)+n2(γ1−1)\gamma = \dfrac{{{n_1}{\gamma _1}({\gamma _2} - 1) + {n_2}{\gamma _2}({\gamma _1} - 1)}}{{{n_1}({\gamma _2} - 1) + {n_2}({\gamma _1} - 1)}}
As we know oxygen is diatomic therefore γ1{\gamma _1} = 1.4
And for carbon dioxide γ2{\gamma _2} = 1.28
Putting the values and solving
2.(1.4(1.28−1))+3(1.28(1.4−1))2(1.28−1)+3(1.4−1)\dfrac{{2.(1.4(1.28 - 1)) + 3(1.28(1.4 - 1))}}{{2(1.28 - 1) + 3(1.4 - 1)}}
On further solving we get
γ=1.33\gamma = 1.33
Hence the option A is correct.

Note :
The heat capacity ratio is important for its applications in thermodynamically reversible processes, especially involving ideal gases; the speed of sound depends on that factor. For an ideal gas. The heat capacity is constant with temperature.