Question 791: Thermodynamics & KTG

Question

The r.m.s. velocity of gas molecules kept at temperature $27^{\circ} \mathrm{C}$ in a vessel is $61 \mathrm{~m} / \mathrm{s}$. Molecular weight of gas is nearly
$$\left[\mathrm{R}=8.31 \frac{\mathrm{~J}}{\mathrm{~mol} \mathrm{~K}}\right]$$

Accepted Answer

The root mean square (r.m.s.) velocity of gas molecules is given by the equation:
$$ v_{	ext{rms}} = \sqrt{\frac{3RT}{M}} $$
where $ v_{	ext{rms}} $ is the r.m.s. velocity, $ R $ is the universal gas constant, $ T $ is the temperature in Kelvin, and $ M $ is the molar mass of the gas in kg/mol.
Given:
$ v_{	ext{rms}} = 61 \ 	ext{m/s} $
$ T = 27^{\circ} 	ext{C} = 27 + 273 = 300 \ 	ext{K} $
$ R = 8.31 \ 	ext{J/(mol \cdot K)} $
To find the molar mass $ M $, rearrange the equation:
$$ M = \frac{3RT}{v_{	ext{rms}}^2} $$
Substitute the given values into the equation:
$$ M = \frac{3 	imes 8.31 \ 	ext{J/(mol} \cdot 	ext{K)} 	imes 300 \ 	ext{K}}{(61 \ 	ext{m/s})^2} $$
Calculating:
Square the velocity:
$$ 61^2 = 3721 \ 	ext{m}^2/	ext{s}^2 $$
Multiply the constants:
$$ 3 	imes 8.31 	imes 300 = 7479 \ 	ext{J/mol} $$
Find $ M $:
$$ M = \frac{7479}{3721} \ 	ext{kg/mol} $$
Calculate the value:
$$ M \approx 2.0105 \ 	ext{kg/mol} $$
Since the molecular weight should be expressed in terms of grams per mole for standard usage:
$$ M \approx 2.0105 	imes 1000 \ 	ext{g/mol} = 2.0105 \ 	ext{g/mol} $$
Therefore, the molecular weight of the gas is nearly 2. The correct option is:
Option A: 2

Answer

2

Answer

4

Answer

28

Answer

32

Thermodynamics & KTG

The r.m.s. velocity of gas molecules kept at temperature 27^{\circ} \mathrm{C} in a vessel is 61 \mathrm{~m} / \mathrm{s}. Molecular weight of gas is nearly $\left[\mathrm{R}=8.31 \frac{\mathrm{~J}}{\mathrm{~mol} \mathrm{~K}}\right]

Choose the correct answer: