Question 67: Thermodynamics & KTG

Question

The molecules of a given mass of a gas have root mean square speed of $$120 \mathrm{~m} / \mathrm{s}$$ at $$88^{\circ} \mathrm{C}$$ and 1 atmospheric pressure. The root mean square speed of the molecules at $$127^{\circ} \mathrm{C}$$ and 2 atmospheric pressure is

Accepted Answer

To find the root mean square speed ($v_{rms}$) of the molecules at a different temperature and pressure, we can use the relations from the kinetic theory of gases. For a given mass of gas, the root mean square speed is directly proportional to the square root of the absolute temperature ($T$) and inversely proportional to the square root of the pressure ($P$), keeping in mind that the amount of gas does not change. Mathematically, this relationship can be expressed as:
$v_{rms} \propto \sqrt{\frac{T}{P}}$
Given that the initial $v_{rms}$ is $120 \, 	ext{m/s}$ at $88^\circ C$ (or $361 K$, since $T_{K} = T_{^\circ C} + 273$) and $1$ atmospheric pressure, and we want to find the $v_{rms}$ at $127^\circ C$ (or $400 K$) and $2$ atmospheric pressures, we can set up the following proportion :
$\frac{v_{rms,1}}{v_{rms,2}} = \sqrt{\frac{T_1/P_1}{T_2/P_2}}$
Using the given data:
$120 / v_{rms,2} = \sqrt{\frac{361 / 1}{400 / 2}}$
$v_{rms,2} = 120 	imes \sqrt{\frac{400 / 2}{361 / 1}}$
$v_{rms,2} = 120 	imes \sqrt{\frac{400}{2 	imes 361}}$
$v_{rms,2} = 120 	imes \sqrt{\frac{400}{722}}$
$v_{rms,2} = 120 	imes \sqrt{\frac{2}{1.805}}$
$v_{rms,2} = 120 	imes \sqrt{1.107694}$
$v_{rms,2} \approx 120 	imes 1.052$
$v_{rms,2} \approx 126.24 \, 	ext{m/s}$
Hence, the correct answer is
Option C: 126.3 m/s
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Answer

105.2 m/s

Answer

1.443 m/s

Answer

126.3 m/s

Answer

88/127 m/s

Thermodynamics & KTG

The molecules of a given mass of a gas have root mean square speed of $120 \mathrm{~m} / \mathrm{s} at 88^{\circ} \mathrm{C} and 1 atmospheric pressure. The root mean square speed of the molecules at 127^{\circ} \mathrm{C}$ and 2 atmospheric pressure is

Choose the correct answer: