For $2 mol$ of ${CO}_{2}$ gas at $300 K$

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Slope of $(\log P)$ vs $(\log V)$ curve is log $(600 R)$ .

Ratio of rotational to vibrational kinetic energy is (all degrees of freedom are activated) $1 : 2$

Translational kinetic energy is $1800 cal$

Ratio of $U_{rms}$ to $U_{mps}$ is $3 : 2$

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Detailed Solution

For ${CO}_{2}$

translational degrees of freedom $= 3$

Rotational degrees of freedom $= 2$

Vibrational degrees of freedom $= 1$

$Energy = \frac{1}{2} nRT$

1)Translational $K . E = 3 \times \frac{1}{2} \times 2 \times 2 \times 300 \Rightarrow 1800 cal given : R = 2 (calories); T = 300; n = 2$

2)Rotational $K . E = 2 \times \frac{1}{2} \times 2 \times 2 \times 300 = 1200 cal$

Vibrational $K . E$ $= 1 \times \frac{1}{2} \times 2 \times 2 \times 300 = 600 cal$

Ratio of rotational /vibrational = $2 : 1$

3) $\frac{U_{rms}}{U_{mps}} = \sqrt{\frac{3}{2}}$

4) slope is $' - 1'$

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