Courses
Courses for Kids
Free study material
Offline Centres
More
Store

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

Last updated date: 12th Aug 2024
Total views: 382.5k
Views today: 3.82k
Verified
382.5k+ views
Hint :Here we are asked to calculate the Heat capacity ratio of two gases i.e. oxygen and carbon dioxide. ${C_p}$ is heat capacity at constant pressure and ${C_v}$ 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 $\gamma$ .

Let ${n_1}$ be the moles of oxygen
${n_2}$ be the moles of carbon dioxide
Total moles = ${n_1} + {n_2} = n$
${\gamma _1},{C_{p1}},{C_{v1}}$ be for oxygen and
${\gamma _2},{C_{p2}},{C_{v2}}$ be for carbon dioxide
Now we have to calculate the value of ${C_p}$ and ${C_v}$ for a given mixture of gases.
${C_v}$ for mixture of gases.
${C_v} = \dfrac{{{C_{v1}}{n_1} + {C_{v2}}{n_2}}}{{{n_1} + {n_2}}}$ ---(1)
And ${C_p}$ for mixture of gases
${C_p} = \dfrac{{{C_{p1}}{n_1} + {C_{p2}}{n_2}}}{{{n_1} + {n_2}}}$ ----(2)
Now we have
$\gamma = \dfrac{{{C_p}}}{{{C_v}}}$ and ${C_v} = \dfrac{R}{{\gamma - 1}}$
Therefore ${C_p} = \dfrac{{\gamma R}}{{\gamma - 1}}$
Now putting all values in equation 1 and 2 we get
${C_p} = \dfrac{{{n_1}{C_{p1}} + {n_2}{C_{p2}}}}{n}$
${C_p} = \dfrac{{{n_1}\dfrac{{{\gamma _1}R}}{{{\gamma _1} - 1}} + {n_2}\dfrac{{{\gamma _2}R}}{{{\gamma _2} - 1}}}}{n}$ ……… (3)
${C_v} = \dfrac{{{n_1}{C_{v1}} + {n_2}{C_{v2}}}}{n}$
${C_v} = \dfrac{{{n_1}\dfrac{R}{{{\gamma _1} - 1}} + {n_2}\dfrac{R}{{{\gamma _2} - 1}}}}{n}$ ………. (4)
From equation 3 and 4 we get
$\gamma = \dfrac{{{C_p}}}{{{C_v}}}$ = $= \dfrac{{\dfrac{{{n_1}\dfrac{{{\gamma _1}R}}{{{\gamma _1} - 1}} + {n_2}\dfrac{{{\gamma _2}R}}{{{\gamma _2} - 1}}}}{n}}}{{\dfrac{{{n_1}\dfrac{R}{{{\gamma _1} - 1}} + {n_2}\dfrac{R}{{{\gamma _2} - 1}}}}{n}}}$
Finally we get
$\gamma = \dfrac{{{n_1}{\gamma _1}({\gamma _2} - 1) + {n_2}{\gamma _2}({\gamma _1} - 1)}}{{{n_1}({\gamma _2} - 1) + {n_2}({\gamma _1} - 1)}}$
As we know oxygen is diatomic therefore ${\gamma _1}$ = 1.4
And for carbon dioxide ${\gamma _2}$ = 1.28
Putting the values and solving
$\dfrac{{2.(1.4(1.28 - 1)) + 3(1.28(1.4 - 1))}}{{2(1.28 - 1) + 3(1.4 - 1)}}$
On further solving we get
$\gamma = 1.33$
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.