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$The ratio of the adiabatic exponent $\gamma$ for a mixture of 1 mole of a monatomic gas and 2 moles of a diatomic gas is$

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Correct answer: A — $$19/13$$

Solution

The adiabatic exponent of a mixture is given by

γ_{mix} = \frac{n _{1} C _{p 1} + n _{2} C _{p 2}}{n _{1} C _{v 1} + n _{2} C _{v 2}}

. For monatomic gas,

C_{v} = 3 R /2

and

C_{p} = 5 R /2

. For diatomic gas,

C_{v} = 5 R /2

and

C_{p} = 7 R /2

. Calculating with

n_{1} = 1

and

n_{2} = 2

gives

19/13