Question

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

Answer 1

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).

Complete answer:
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”.

Additional Information:
According to the kinetic theory of gases, particles are in a state of random motion. Particles move at a different speed and collide and change direction. We use the term speed to represent the movement of particles. We cannot measure the speed of the individual particle, so we often use the term’ average’. Particles can have opposite directions. Since gas particles are random motion we cannot say about how many particles are moving in which direction. Thus the average speed is always taken as zero considering that for every moving particle there is one which moves in the opposite direction.
Thus we use the alternate method to get the velocities which are by squaring the velocities and taking the square root. Since these values are free from the direction we referred to it as average speed. The root means square velocity for the particle is the square root of average speed squared of molecules in gas.
${{\text{V}}_{\text{rms}}}\text{=}\sqrt{\dfrac{\text{3RT}}{\text{M}}}$
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.

Note: The root means square speed considers both molecular weight (M) and temperature (T). Two of these factors are directly related to the kinetic energy of gas particles. Students should not get confused with the velocity and speed. The RMS calculation gives the root mean square speed. It does not give us velocity. Velocity is a vector quantity that has both direction and magnitude. But RMS calculation only gives the magnitude or speed of the particles.