Two moles of an ideal gas
Question:

Two moles of an ideal gas with $\frac{C_{p}}{C_{V}}=\frac{5}{3}$ are mixed with 3 

moles of another ideal gas with $\frac{C_{p}}{C_{V}}=\frac{4}{3}$. The value of $\frac{C_{p}}{C_{V}}$ for the mixture is:

  1. (1) $1.45$

  2. (2) $1.50$

  3. (3) $1.47$

  4. (4) $1.42$


Correct Option: , 4

Solution:

(4) Using, $\gamma_{\text {mixture }}=\frac{n_{1} C_{p_{1}}+n_{2} C_{p_{2}}}{n_{1} C_{v_{1}}+n_{2} C_{v_{2}}}$

$\Rightarrow \frac{n_{1}}{\gamma_{1}-1}+\frac{n_{2}}{\gamma_{2}-1}=\frac{n_{1}+n_{2}}{\gamma_{m}-1}$

$\Rightarrow \frac{3}{\frac{4}{3}-1}+\frac{2}{\frac{5}{3}-1}=\frac{5}{\gamma_{m}-1}$

$\Rightarrow \frac{9}{1}+\frac{2 \times 3}{2}=\frac{5}{\gamma_{m}-1}$

$\Rightarrow \gamma_{m}-1=\frac{5}{12}$

$\Rightarrow \gamma_{m}=\frac{17}{12}=1.42$

Administrator

Leave a comment

Please enter comment.
Please enter your name.