An adiabatic vessel contains n₁ moles of diatomic gas. Moment of inertia of each molecule is $I = 2.76 \times 10^{- 46} k g m^{2}$ and root mean square angular velocity is $ω_{0} = 5 \times 10^{12} r a d s^{- 1}$ . Another adiabatic vessel contains n₂= 5 moles of a monoatomic gas at a temperature 470K. Assume gases to be ideal, calculate root-mean square angular velocity of diatomic molecules (in multiple of $10^{12} r a d s^{- 1}$ ) when the two vessels are connected by a thin tube of negligible volume.

Very difficult

Solution

By energy conservation,
$(f_{1} n_{1} + f_{2} n_{2}) T_{f} = f_{1} n_{1} T_{1} + f_{2} n_{2} T_{2}$ (1)
Here, $f_{1} = 5, n_{1} = 3, T_{1} = 250 K$ and
$f_{2} = 3, n_{2} = 5, T_{2} = 470 K$
Substituting in (1),
$T_{f} = \frac{(5 \times 3) \times 250 + (3 \times 5) \times 470}{(5 \times 3) + (3 \times 5)} = 360 K$
If RMS angular velocity of diatomic gas molecules is ${(ω_{R M S})}_{f}$ , then according to law of equipartition of energy,
$\frac{1}{2} I {(ω_{R M S})}_{f}^{2} = k T$
Or,
$\Rightarrow {(ω_{R M S})}_{f} = \sqrt{\frac{2 k T}{I}} = \sqrt{\frac{2 \times 1.38 \times 10^{- 23} \times 360}{2.76 \times 10^{- 46}}} = 6 \times 10^{12} r a d / s$ .

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