If two moles of diatomic gas and one mole of mono atomic gas are mixed then the ratio of specific heats is -

\frac{7}{3}

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\frac{5}{4}

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\frac{19}{13}

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\frac{15}{19}

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Solution

The correct option is C $\frac{19}{13}$
$C_{p_{e q}} = \frac{n_{1} C_{p_{1}} + n_{2} C_{p_{2}}}{n_{1} + n_{2}}$ $C_{v_{e q}} = \frac{n_{1} C_{v_{1}} + n_{2} C_{v_{2}}}{n_{1} + n_{2}}$
$r_{e q} = \frac{C_{P_{e q}}}{C_{v_{e q}}} = \frac{2 \times \frac{7}{2} + 1 \times \frac{5}{2}}{2 \times \frac{5}{2} + 1 \times \frac{3}{2}}$

$= \frac{19}{13}$

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If two moles of diatomic gas and one mole of mono atomic gas are mixed then the ratio of specific heats is -

The correct option is C 1913Cpeq=n1Cp1+n2Cp2n1+n2 Cveq=n1Cv1+n2Cv2n1+n2 req=CPeqCveq=2×72+1×522×52+1×32 =1913