First slide

Heat given

Question

A closed container contains 2 moles of hydrogen gas. If 50% of hydrogen molecules are dissociated into atomic hydrogen, the molar specific at constant volume of the mixture will be

Easy

A

$\frac{9 R}{2}$
B

2R
C

$\frac{7 R}{6}$
D

$\frac{11 R}{6}$

Solution

(4)
After dissociation, the mixture contains 1 mole of diatomic molecules and 2 moles of mono-atomic molecules
$∴ C_{V} = \frac{n_{1} . {(C_{V})}_{1} + n_{2} {(C_{V})}_{2}}{n_{1} + n_{2}} = \frac{1. \frac{5 R}{2} + 2. \frac{3 R}{2}}{1 + 2} = \frac{11 R}{6}$

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