Question

4 moles of an ideal gas having $\gamma$ = 1.67 are mixed with 2 moles of another ideal gas having $\gamma$ = 1.4. Find the equivalent value of $\gamma$ for the mixture.

• A. 1.4
• B. 1.33
• C. 1.54
• D. 0.78

Answer 1

The correct option is C 1.54

Let
CV´= molar heat capacity of the first gas,
CV´´ = molar heat capacity of the second gas,
CV = molar heat capacity of the mixture,
and similar symbols for other quantities. Then,
$\gamma =\frac{C^{\prime }p}{C_V^{\prime }}=1.67$
and
$C^{\prime }P=C^{\prime }V+R$
This gives
$C^{\prime }V=\frac{3}{2}R$ and $C_P^{\prime }=\frac{5}{2}R$
Similarly, $\gamma$ = 1.4 gives $C_V^{\prime \prime }=\frac{5}{2}R$
and $C_P^{\prime \prime }=\frac{7}{2}R$
Suppose the temperature of the mixture is increased by dT. The increase in the internal energy of the first gas = $n_1{C}_V^{\prime }dT.$
The increase in internal energy of the second gas = $n_2{C}_V^{\prime \prime }dT.$
Thus,
$(n_1+n_2)C_{V}dT=n_1{C}_V^{\prime }dT+n_{2}C{V}^{\prime \prime }dTor,C_V=\frac{n_1{C}_V^{\prime }+n_2{C}_V^{\prime \prime }}{n_1+n_2}...\left(1\right)C_P=C_V+R=\frac{n_1{C}_V^{\prime }+n_2{C}_V^{\prime \prime }}{n_1+n_2}+R=\frac{n_1(C_V^{\prime }+R)+n_2(C_V^{\prime \prime }+R)}{n_1+n_2}=\frac{n_1{C}_p^{\prime }+n_2{C}_p^{\prime \prime }}{n_1+n_2}...\left(2\right)\text{From (1) and (2),}\gamma =\frac{C_p}{C_V}=\frac{n_1{C}_p^{\prime }+n_2{C}_p^{\prime \prime }}{n_1{C}_V^{\prime }+n_2{C}_V^{\prime \prime }}$
$\frac{4\times \frac{5}{2}R+2\times \frac{7}{2}R}{4\times \frac{3}{2}R+2\times \frac{5}{2}R}=1.54$