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Question: If two moles of diatomic gas and one mole of monoatomic gas are mixed with then the ratio of specifi...

If two moles of diatomic gas and one mole of monoatomic gas are mixed with then the ratio of specific heats is

A

73\frac{7}{3}

B

54\frac{5}{4}

C

1913\frac{19}{13}

D

1519\frac{15}{19}

Answer

1913\frac{19}{13}

Explanation

Solution

μ1=1\mu_{1} = 1, γ1=53\gamma_{1} = \frac{5}{3} (for monoatomic gas) and μ2=2\mu_{2} = 2,

γ2=75\gamma_{2} = \frac{7}{5} (for diatomic gas)

From formula γmixture=μ1γ1γ1−1+μ2γ2γ2−1μ1γ1−1+μ2γ2−1=1×5353−1+2×7575−1153−1+275−1=5/2+73/2+5=1913\gamma_{\text{mixture}} = \frac{\frac{\mu_{1}\gamma_{1}}{\gamma_{1} - 1} + \frac{\mu_{2}\gamma_{2}}{\gamma_{2} - 1}}{\frac{\mu_{1}}{\gamma_{1} - 1} + \frac{\mu_{2}}{\gamma_{2} - 1}} = \frac{\frac{1 \times \frac{5}{3}}{\frac{5}{3} - 1} + \frac{2 \times \frac{7}{5}}{\frac{7}{5} - 1}}{\frac{1}{\frac{5}{3} - 1} + \frac{2}{\frac{7}{5} - 1}} = \frac{5/2 + 7}{3/2 + 5} = \frac{19}{13}