Courses
Courses for Kids
Free study material
Offline Centres
More
Store

# The root mean square speed of the hydrogen molecule of an ideal hydrogen gas kept in a gas chamber at ${{0}^{0}}$C is $\text{3180 metre/second}$. The pressure on the hydrogen gas is: (density of hydrogen gas is $8.99\times {{10}^{-2}}kg/{{m}^{3}}$,$\text{1 atm=1}\text{.01 }\!\!\times\!\!\text{ 1}{{\text{0}}^{\text{5}}}\text{N/}{{\text{m}}^{\text{2}}}$). A) 1.0 atmosphere B) 1.5 atmosphere C) 2.0 atmosphere D) 3.0 atmosphere

Last updated date: 20th Jun 2024
Total views: 413.1k
Views today: 6.13k
Verified
413.1k+ views
Hint: The root mean square speed denoted by ${{\text{V}}_{\text{rms}}}$ is directly proportional to the pressure on the gas P and inversely related to the density of gas in the chamber (d).

Root mean square speed ${{\text{V}}_{\text{rms}}}$ is the square root of the average of the square of the speed. The RMS value is used for the gas sample as the net speed is zero but the particles are moving in all directions.

-We are given with the following data;
The root mean square speed of ${{\text{H}}_{\text{2}}}$ at ${{0}^{0}}\text{c}$ is $\text{3180m/s}$
The density of hydrogen gas kept in the chamber is $\text{8}\text{.99 }\!\!\times\!\!\text{ 1}{{\text{0}}^{\text{-2}}}\text{Kg/}{{\text{m}}^{\text{3}}}$
$\text{1 atm=1}\text{.01 }\!\!\times\!\!\text{ 1}{{\text{0}}^{\text{5}}}\text{N/}{{\text{m}}^{\text{2}}}$
-The root means square speed is related to the density of ${{\text{H}}_{\text{2}}}$ gas kept in the chamber and pressure on the gas. The relation between ${{\text{V}}_{\text{rms}}}$, P and d is given as,
${{\text{V}}_{\text{rms}}}\text{=}\sqrt{\dfrac{\text{3P}}{\text{d}}}$----------- (1)

Where,
${{\text{V}}_{\text{rms}}}$=root mean square speed
P=pressure on ${{\text{H}}_{\text{2}}}$ gas
d=density of ${{\text{H}}_{\text{2}}}$ gas
-Let's substitute the values given in equation (1). We get,
${{\text{V}}_{\text{rms}}}\text{=}\sqrt{\dfrac{\text{3P}}{\text{d}}}$
⟹$\text{3180m/s=}\sqrt{\dfrac{\text{3P}}{\text{8}\text{.99 }\!\!\times\!\!\text{ 1}{{\text{0}}^{\text{-2}}}\text{Kg/}{{\text{m}}^{\text{3}}}}}$
On squaring on both sides. We get,
⟹${{\text{(3180m/s)}}^{2}}\text{=}{{\left( \sqrt{\dfrac{\text{3P}}{\text{8}\text{.99 }\!\!\times\!\!\text{ 1}{{\text{0}}^{\text{-2}}}\text{Kg/}{{\text{m}}^{\text{3}}}}} \right)}^{2}}$
⟹${{\text{(3180m/s)}}^{2}}\text{=}\dfrac{\text{3P}}{\text{8}\text{.99 }\!\!\times\!\!\text{ 1}{{\text{0}}^{\text{-2}}}\text{Kg/}{{\text{m}}^{\text{3}}}}$

Rearrange the above equation for pressure P.
⟹$\text{P=}\dfrac{{{\text{(3180m/s)}}^{\text{2}}}\text{ }\!\!\times\!\!\text{ 8}\text{.99 }\!\!\times\!\!\text{ 1}{{\text{0}}^{\text{-2}}}\text{Kg/}{{\text{m}}^{\text{3}}}}{\text{3}}$
⟹$\text{P=}\dfrac{{{\text{(3180)}}^{\text{2}}}\text{ }\!\!\times\!\!\text{ 8}\text{.99 }\!\!\times\!\!\text{ 1}{{\text{0}}^{\text{-2}}}\frac{Kg}{{{m}^{2}}}\dfrac{m}{{{s}^{2}}}}{\text{3}}$
Since we know that $\text{Kg}\dfrac{\text{m}}{{{\text{s}}^{\text{2}}}}\text{=N}$

The above equation become,
⟹$\text{P=}\dfrac{{{\text{(3180)}}^{\text{2}}}\text{ }\!\!\times\!\!\text{ 8}\text{.99 }\!\!\times\!\!\text{ 1}{{\text{0}}^{\text{-2}}}\dfrac{\text{N}}{{{\text{m}}^{\text{2}}}}}{\text{3}}$---------- (2)
We know that, $\text{1 atm=1}\text{.01 }\!\!\times\!\!\text{ 1}{{\text{0}}^{\text{5}}}\text{N/}{{\text{m}}^{\text{2}}}$
Let’s, divide equation (2) by $\text{1}\text{.01 }\!\!\times\!\!\text{ 1}{{\text{0}}^{\text{5}}}\text{N/}{{\text{m}}^{\text{2}}}$. We get,
⟹$\text{P=}\dfrac{\text{9}\text{.09 }\!\!\times\!\!\text{ 1}{{\text{0}}^{\text{5}}}}{\text{3 }\!\!\times\!\!\text{ 1}\text{.01 }\!\!\times\!\!\text{ 1}{{\text{0}}^{\text{5}}}}\text{atm}$
⟹$\text{P=3}\text{.0atm}$

So, the correct answer is “Option D”.

${{\text{V}}_{\text{rms}}}\text{=}\sqrt{\dfrac{\text{3RT}}{\text{M}}}$
Where, ${{\text{V}}_{\text{rms}}}$ is the root mean square of the speed, M is the molecular mass of gas in $\text{Kg/mole}$, R is gas constant and T is the temperature in kelvin.