Question 893: Thermodynamics & KTG

Question

The molecular masses of helium and oxygen are 4 and 32 respectively. The ratio of r.m.s. speed of helium at 327$$^\circ$$ to r.m.s. speed of oxygen at 27$$^\circ$$ will be

Accepted Answer

The root mean square (r.m.s) speed of a gas is given by the equation:
$$ v_{rms} = \sqrt{\dfrac{3kT}{m}} $$
where:
$$v_{rms}$$ is the root mean square speed.
$$k$$ is the Boltzmann constant.
$$T$$ is the temperature in Kelvin.
$$m$$ is the molecular mass in kg.
Let's denote the molecular masses by $$ m_H = 4 \, 	ext{u} $$ for Helium and $$ m_O = 32 \, 	ext{u} $$ for Oxygen. Also, convert the temperatures from Celsius to Kelvin:
$$ T_1 = 327^\circ 	ext{C} + 273 = 600 \, 	ext{K} $$
$$ T_2 = 27^\circ 	ext{C} + 273 = 300 \, 	ext{K} $$
We need to find the ratio of r.m.s. speed of helium at 327$$^\circ$$C to the r.m.s. speed of oxygen at 27$$^\circ$$C:
$$ \frac{v_{rms, He}}{v_{rms, O_2}} $$
Using the r.m.s. speed formula, we have:
$$ v_{rms, He} = \sqrt{\dfrac{3k \cdot 600}{4}} $$
$$ v_{rms, O_2} = \sqrt{\dfrac{3k \cdot 300}{32}} $$
We take the ratio:
$$ \frac{v_{rms, He}}{v_{rms, O_2}} = \frac{\sqrt{\dfrac{3k \cdot 600}{4}}}{\sqrt{\dfrac{3k \cdot 300}{32}}} $$
Simplify the expression:
$$ \frac{v_{rms, He}}{v_{rms, O_2}} = \sqrt{\dfrac{600}{4}} \, \div \, \sqrt{\dfrac{300}{32}} = \dfrac{\sqrt{600/4}}{\sqrt{300/32}} $$
Further simplify this expression:
$$ \frac{v_{rms, He}}{v_{rms, O_2}} = \sqrt{\dfrac{600}{4} \, 	imes \, \dfrac{32}{300}} = \sqrt{\dfrac{600 \cdot 32}{4 \cdot 300}} = \sqrt{\dfrac{19200}{1200}} = \sqrt{16} = 4 $$
Thus, the ratio of the r.m.s. speed of helium to the r.m.s. speed of oxygen is 4:1. The correct option is:
Option D: 4 : 1

Answer

1 : 6

Answer

8 : 1

Answer

1 : 8

Answer

4 : 1

Thermodynamics & KTG

The molecular masses of helium and oxygen are 4 and 32 respectively. The ratio of r.m.s. speed of helium at 327$^\circ to r.m.s. speed of oxygen at 27^\circ$ will be

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