Question 214: Thermodynamics & KTG

Question

Three vessels of equal volume contain gases at the same temperature and pressure. The first vessel contains neon (monoatomic), the second contains chlorine (diatomic) and third contains uranium hexafloride (polyatomic). Arrange these on the basis of their root mean square speed $$\left(v_{\mathrm{rms}}\right)$$ and choose the correct answer from the options given below:

Accepted Answer

The root mean square speed ($$v_{rms}$$) of a gas is given by the formula:
$$v_{rms} = \sqrt{\frac{3kT}{m}}$$
where $$k$$ is the Boltzmann constant, $$T$$ is the temperature, and $$m$$ is the molar mass of the gas molecules.
The vessels contain neon (monoatomic), chlorine (diatomic), and uranium hexafluoride (polyatomic) gases. Their molar masses are:
Neon: $$20.18\,	ext{g/mol}$$ (monoatomic)
Chlorine: $$2 	imes 35.45 = 70.90\,	ext{g/mol}$$ (diatomic)
Uranium hexafluoride: $$238.03 + 6 	imes 18.998 = 352.03\,	ext{g/mol}$$ (polyatomic)
Since the temperature and the Boltzmann constant are the same for all gases, the root mean square speed is inversely proportional to the square root of the molar mass:
$$v_{rms} \propto \frac{1}{\sqrt{m}}$$
The lighter the gas, the higher its root mean square speed. Comparing the molar masses of the gases, we find that neon is the lightest, followed by chlorine, and uranium hexafluoride is the heaviest. Therefore, the root mean square speeds will be:
$$v{rms}(	ext{mono}) > v{rms}(	ext{dia}) > v_{rms}(	ext{poly})$$

Answer

$$\mathrm{v}_{\mathrm{rms}}($$ mono $$)=\mathrm{v}_{\mathrm{rms}}($$ dia $$)=\mathrm{v}_{\mathrm{rms}}($$ poly $$)$$

Answer

$$\mathrm{v}_{\mathrm{rms}}$$ (mono) $$ > \mathrm{v}_{\mathrm{rms}}($$ dia $$) > \mathrm{v}_{\mathrm{rms}}$$ (poly)

Answer

$$\mathrm{v}_{\mathrm{rms}}$$ (dia) $$ < \mathrm{v}_{\mathrm{rms}}$$ (poly) $$ < \mathrm{v}_{	ext {rms }}$$ (mono)

Answer

$$\mathrm{v}_{\mathrm{rms}}$$ (mono) $$ < \mathrm{v}_{\mathrm{rms}}$$ (dia) $$ < \mathrm{v}_{\mathrm{rms}}$$ (poly)

Thermodynamics & KTG

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