Question

Two moles of an ideal gas with $\frac{C_p}{C_v}=5/3$ are mixed with 3 moles of another ideal gas with $\frac{C_p}{C_v}=4/3$. The value of $\frac{C_p}{C_v}$ for the mixture is

• A. $1.47$
• B. $1.42$
• C. $1.45$
• D. $1.50$

Answer 1

The correct option is B $1.42$

$$\begin{array}{cc}& \text{For first gas having}\gamma =\frac{C_p}{C_v}=\frac{5}{3}\\ & \text{Using formula}{C}_p=\frac{R\gamma }{\gamma -1}\\ & C_v=\frac{R}{\gamma -1}\\ & C_p=\frac{5R}{2}{C}_v=\frac{3R}{2}\\ & \text{Similarly for}{2}^{nd}\text{gas having}\gamma =\frac{C_p}{C_v}=\frac{4}{3}\\ & C_p=4R{C}_v=3R\\ & \text{Now}\gamma \text{of mixture}=\frac{n_1{C}_{p_1}+n_2{C}_{p_2}}{n_1{C}_{v_1}+n_2{C}_{v_2}}\\ & \text{Given that}{n}_1=2\text{and}{n}_2=3\\ & \gamma =\frac{2\times \frac{5R}{2}+3\times 4R}{2\times \frac{3R}{2}+3\times 3R}=\frac{17}{12}=1.42\end{array}$$