Question

A gaseous mixture consists of 16g of helium and 16g of oxygen. The ratio $\dfrac{{{C_P}}}{{{C_V}}}$ of mixture is :
(A). 1.4
(B). 1.54
(C). 1.59
(D). 1.62

Answer 1

Hint: - The ratio of the specific heats $\gamma = \dfrac{{{C_P}}}{{{C_V}}}$is a factor in adiabatic engine processes and it is very helpful in determining the speed of sound in a gas.
Formula Used: - ${C_v} = \dfrac{{{n_1}{C_{V1}} + {n_2}{C_{V2}}}}{{{n_1} + {n_2}}}$
$\gamma = \dfrac{{{C_P}}}{{{C_V}}}$

Complete step-by-step solution -
It is the ratio of the heat capacity at constant pressure ${C_P}$ to heat capacity at constant volume ${C_V}$. Sometimes it is also known as the isentropic expansion factor and is denoted by $\gamma$ which is called gamma.
Therefore,
\gamma = \dfrac{{{C_P}}}{{{C_V}}} = \dfrac{{\mathop {{C_P}}\limits^\\_ }}{{\mathop {{C_V}}\limits^\\_ }} = \dfrac{{{c_p}}}{{{c_v}}}
This ratio is $\gamma$ = 1.66 for an ideal monatomic gas and $\gamma$ = 1.4 for air, which is predominantly a diatomic gas.

${C_V}$ is defined as the amount of heat required to raise the temperature of one mole of gas by $1^\circ C$ at constant volume. ${C_V}$ can be written as, ${C_V}$=$\dfrac{1}{2}fR$
Where $f$ is the degree of freedom of the gas molecule and $R$ is gas constant.
${C_P}$ is the amount of heat required to raise the temperature of one mole of gas by $1^\circ C$ at constant pressure.
Therefore, $\gamma = \dfrac{{{C_P}}}{{{C_V}}} = 1 + \dfrac{2}{f}$, here $f$ is the degree of freedom of the gas molecules.
As $\gamma$ gamma is a ratio of two quantities having the same unit therefore, $\gamma$ is a dimensionless quantity.
${M_{He}} =$4
${m_{He}} =$16 g
${M_o} =$32
${m_o} =$16g
Specific heat of mixture at constant volume is given by,
${C_v} = \dfrac{{{n_1}{C_{V1}} + {n_2}{C_{V2}}}}{{{n_1} + {n_2}}}$----(i)
For helium gas,
No of moles, ${n_1} = \dfrac{{{m_{He}}}}{{{M_{He}}}}$
${n_1} = \dfrac{{16}}{4}$
Therefore, ${\gamma _1} = \dfrac{5}{3}$ = 1.66
For Oxygen gas,
No of moles, ${n_2} = \dfrac{{{m_o}}}{{{M_o}}}$
${n_2} = \dfrac{{16}}{{32}} = \dfrac{1}{2}$
Therefore, ${\gamma _2} = \dfrac{7}{5}$ = 1.4
At constant volume specific heat of helium
${C_{V1}} = \dfrac{R}{{{\gamma _1} - 1}}$=$\dfrac{R}{{\dfrac{5}{3} - 1}} = \dfrac{3}{2}R$
At constant volume specific heat of oxygen
${C_{V2}} = \dfrac{R}{{{\gamma _2} - 1}}$=$\dfrac{R}{{\dfrac{7}{5} - 1}} = \dfrac{5}{2}R$
Putting these values in equation (i) we get
${C_v} = \dfrac{{4 \times \dfrac{3}{2}R + \dfrac{1}{2} \times \dfrac{5}{2}R}}{{\dfrac{4}{1} + \dfrac{1}{2}}}$
${C_v} = \dfrac{{29}}{{18}}R = \dfrac{R}{{\gamma - 1}}$
$\gamma = \dfrac{{18}}{{29}} + 1$
$\dfrac{{{C_P}}}{{{C_V}}} = \gamma = 1.62$
Therefore, our answer is option D

Note: - The heat capacity ratio which is also known as the adiabatic index, is the ratio of specific heats or Laplace's coefficient. We must remember this equation and the properties of gas for solving these types of questions.