# 1 mole of a gas with $\dpi{100} \gamma =7/5$ is mixed with 1 mole of a gas with $\dpi{100} \gamma =5/3$ , then the value of $\dpi{100} \gamma$  for the resulting mixture is Option 1) $7/5\;$ Option 2) $\; 2/5\;$ Option 3) $\; 24/16\;$ Option 4) $\; 12/7$

As we learnt in

First law in isobaric process -

$\Delta U= n\, C_{v}\Delta T$

$= n\frac{R}{\gamma -1}\Delta T$

- wherein

$\Delta Q= \Delta U+W$

$= n\frac{\gamma \: R}{\gamma -1}\cdot \Delta T$

$\Delta Q= nC_{p}\: \Delta T$

$\gamma 1 =\frac{7}{5}\rightarrow CV_{1}= \frac{R}{\gamma 1-1}= \frac{5R}{2},C_{P1}=\frac{7R}{2}$

$\gamma 2 =\frac{5}{3}\rightarrow CV_{2}= \frac{R}{\gamma 2-1}= \frac{3R}{2},C_{P2}=\frac{5R}{2}$

$\gamma^{mixture}=\frac{C_{p}^{mix}}{C_{v}^{mix}}=\frac{n_{1}C_{p1}+n_{2}C_{p2}}{n_{1}C_{v1}+n_{2}C_{v2}}= \frac{6R}{4R}=1.5$

Option 1)

$7/5\;$

Option 2)

$\; 2/5\;$

Option 3)

$\; 24/16\;$

Option 4)

$\; 12/7$

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