
If the root means square velocity is \[{\left( {{\rm{3}}{{\rm{0}}^{\rm{2}}}{\rm{R}}} \right)^{\frac{{\rm{1}}}{{\rm{2}}}}}\] at \[{\rm{27^\circ C}}\] then calculate the molar mass of gas in kilograms.
A. 1
B. 2
C. 4
D. 0.001
Answer
161.4k+ views
Hint: Root mean square velocity or r.m.s velocity is interpreted as the square root of the mean of the squares of the distinct velocities.
It is signified by \[{{\rm{V}}_{{\rm{rms}}}}\].
Formula Used:
\[{{\rm{V}}_{{\rm{rms}}}}{\rm{ = }}\sqrt {\frac{{{\rm{3RT}}}}{{\rm{M}}}} \] where
R = Universal gas constant
T = Temperature
M = Gram molecular mass of the gas molecule
Complete Step by Step Solution:
A gas comprises molecules present at large distances.
The molecules are independent to change positions in the area in which it is enclosed.
These molecules have continual motion in varied directions at varied velocities.
They keep on striking each other and the walls of the container which is defined as a molecular collision.
The expression for r.m.s velocity is given by the expression:
Given
\[{{\rm{V}}_{{\rm{rms}}}}{\rm{ = }}\sqrt {\frac{{{\rm{3RT}}}}{{\rm{M}}}} \]
\[{{\rm{V}}_{{\rm{rms}}}}{\rm{ = }}{\left( {{\rm{3}}{{\rm{0}}^{\rm{2}}}{\rm{R}}} \right)^{\frac{{\rm{1}}}{{\rm{2}}}}}\]
For calculation of rms velocity, the temperature must be in Kelvin.
So, temperature=27°C=(27+273)=300K
\[ \Rightarrow \sqrt {{{\left( {{\rm{30}}} \right)}^{\rm{2}}}{\rm{R}}} {\rm{ = }}\sqrt {\frac{{{\rm{3R}}\left( {{\rm{300K}}} \right)}}{{\rm{M}}}} \]
Squaring both sides we get,
\[ \Rightarrow {\left( {{\rm{30}}} \right)^{\rm{2}}}{\rm{ = }}\frac{{{\rm{3}}\left( {{\rm{300K}}} \right)}}{{\rm{M}}}\]
\[ \Rightarrow {\rm{M = 3}}\left( {{\rm{300}}} \right){\rm{30 \times 30g}}\]
So, M=1g=0.001kg
Hence, the molecular mass of the gas is 0.001kg.
So, option D is correct.
Note: r.m.s velocity and the square root of temperature have a direct relationship. Root mean square velocity and the square root of molecular mass are inversely related. While calculating the numerical value of r.m.s velocity, the temperature is to be taken in kelvin and molar mass is to be taken in kg always to get the value of r.m.s velocity in m/s.
It is signified by \[{{\rm{V}}_{{\rm{rms}}}}\].
Formula Used:
\[{{\rm{V}}_{{\rm{rms}}}}{\rm{ = }}\sqrt {\frac{{{\rm{3RT}}}}{{\rm{M}}}} \] where
R = Universal gas constant
T = Temperature
M = Gram molecular mass of the gas molecule
Complete Step by Step Solution:
A gas comprises molecules present at large distances.
The molecules are independent to change positions in the area in which it is enclosed.
These molecules have continual motion in varied directions at varied velocities.
They keep on striking each other and the walls of the container which is defined as a molecular collision.
The expression for r.m.s velocity is given by the expression:
Given
\[{{\rm{V}}_{{\rm{rms}}}}{\rm{ = }}\sqrt {\frac{{{\rm{3RT}}}}{{\rm{M}}}} \]
\[{{\rm{V}}_{{\rm{rms}}}}{\rm{ = }}{\left( {{\rm{3}}{{\rm{0}}^{\rm{2}}}{\rm{R}}} \right)^{\frac{{\rm{1}}}{{\rm{2}}}}}\]
For calculation of rms velocity, the temperature must be in Kelvin.
So, temperature=27°C=(27+273)=300K
\[ \Rightarrow \sqrt {{{\left( {{\rm{30}}} \right)}^{\rm{2}}}{\rm{R}}} {\rm{ = }}\sqrt {\frac{{{\rm{3R}}\left( {{\rm{300K}}} \right)}}{{\rm{M}}}} \]
Squaring both sides we get,
\[ \Rightarrow {\left( {{\rm{30}}} \right)^{\rm{2}}}{\rm{ = }}\frac{{{\rm{3}}\left( {{\rm{300K}}} \right)}}{{\rm{M}}}\]
\[ \Rightarrow {\rm{M = 3}}\left( {{\rm{300}}} \right){\rm{30 \times 30g}}\]
So, M=1g=0.001kg
Hence, the molecular mass of the gas is 0.001kg.
So, option D is correct.
Note: r.m.s velocity and the square root of temperature have a direct relationship. Root mean square velocity and the square root of molecular mass are inversely related. While calculating the numerical value of r.m.s velocity, the temperature is to be taken in kelvin and molar mass is to be taken in kg always to get the value of r.m.s velocity in m/s.
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