Question

An adiabatic vessel contains 3 moles of a diatomic gas. The r.m.s angular velocity is 5 × 1012 rad/ sec . Another adiabatic vessel contains 5 moles of a monatomic gas at a temperature 470 K. Assume gases to be ideal. Calculate rms angular velocity of diatomic molecules (in rad/sec.) when the two vessels are connected by a thin tube of negligible volume. Take $\frac{I}{k}$ = 2 × 10-23 S.I units.

Answer 1

6 × 1012 rad/sec.

Answer 2

The RMS angular velocity ($\omega_{rms}$) is related to temperature ($T$) by $\frac{1}{2} I \omega_{rms}^2 = k_B T$, so $\omega_{rms} = \sqrt{\frac{2k_B T}{I}}$. Given $\frac{I}{k_B} = 2 \times 10^{-23} s^2 \cdot K$, we have $\frac{k_B}{I} = 0.5 \times 10^{23} \frac{1}{s^2 \cdot K}$. Thus, $\omega_{rms} = \sqrt{10^{23} \frac{T}{s^2 \cdot K}}$.

1. Initial Temperature ($T_1$) of Diatomic Gas: Using $\omega_{rms,1} = 5 \times 10^{12}$ rad/sec: $(5 \times 10^{12})^2 = 10^{23} \frac{T_1}{s^2 \cdot K} \implies T_1 = 250$ K.

2. Initial Internal Energies: Diatomic gas ($n_1=3$, $f_1=5$): $U_1 = n_1 \frac{f_1}{2} R T_1 = 3 \times \frac{5}{2} R \times 250 = 1875R$. Monatomic gas ($n_2=5$, $f_2=3$, $T_2=470$ K): $U_2 = n_2 \frac{f_2}{2} R T_2 = 5 \times \frac{3}{2} R \times 470 = 3525R$. Total initial internal energy: $U_{total, initial} = 1875R + 3525R = 5400R$.

3. Final Equilibrium Temperature ($T_f$): Total moles $n_{total} = 3+5=8$. $U_{total, final} = (n_1 \frac{f_1}{2} + n_2 \frac{f_2}{2}) R T_f = (\frac{15}{2} + \frac{15}{2}) R T_f = 15 R T_f$. Equating energies: $5400R = 15 R T_f \implies T_f = 360$ K.

4. Final RMS Angular Velocity: $\omega_{rms, final} = \sqrt{10^{23} \frac{360 K}{s^2 \cdot K}} = \sqrt{36 \times 10^{24} s^{-2}} = 6 \times 10^{12}$ rad/sec.