Question 852: Thermodynamics & KTG

Question

If temperature of gas molecules is raised from $$127^{\circ} \mathrm{C}$$ to $$527^{\circ} \mathrm{C}$$, the ratio of r.m.s. speed of the molecules is respectively

Accepted Answer

We know
$$\mathrm{V}_{\mathrm{rms}}=\sqrt{\frac{3 \mathrm{RT}}{\mathrm{m}}}$$
The temperature of the same gas molecule is raised.
$$\mathrm{V}_{\mathrm{ms}} \propto \sqrt{\mathrm{T}}$$
$$	herefore \quad$$ The ratio of the velocities is
$$\begin{aligned}
& \frac{\mathrm{V}_1}{\mathrm{~V}_2}=\frac{\sqrt{\mathrm{T}_1}}{\sqrt{\mathrm{T}_2}} \
& \frac{\mathrm{V}_1}{\mathrm{~V}_2}=\frac{\sqrt{400}}{\sqrt{800}} \
& \frac{\mathrm{V}_1}{\mathrm{~V}_2}=\frac{1}{\sqrt{2}}
\end{aligned}$$
Convert temperatures given in centigrade to Kelvin before calculation.

Answer

$$1: 2$$

Answer

$$2: 1$$

Answer

$$1: \sqrt{2}$$

Answer

$$2: \sqrt{2}$$

Thermodynamics & KTG

If temperature of gas molecules is raised from $127^{\circ} \mathrm{C} to 527^{\circ} \mathrm{C}$, the ratio of r.m.s. speed of the molecules is respectively

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