For a gas $\frac{R}{c_{v}}=0 \cdot 67$. This gas is made up of molecules which are

For a gas $\frac{R}{c_{v}}=0 \cdot 67$. This gas is made up of molecules which are
  1. diatomic.
  2. polyatomic.
  3. monoatomic.
  4. mixture of diatomic and polyatomic.

Solution

$\frac{R}{C_{v}}=0.67$ $\therefore \mathrm{R}=0.67 \mathrm{C}_{\mathrm{v}}$ $\therefore C_{p}-C_{v}=0.67 \mathrm{C}_{v}$ $\therefore C_{p}=1.67 \mathrm{C}_{v}$ $\therefore \gamma=\frac{\mathrm{C}_{\mathrm{p}}}{\mathrm{C}_{\mathrm{v}}}=1.67$ For a monoatomic gas $\gamma=\frac{5}{3}$ or $1.67$

Asked in: MHT CET 2020 (19 Oct Shift 1)

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