
The r.m.s velocity of a certain gas is v at 300K. The temperature, at which the r.m.s. velocity becomes double:
A. 1200K
B. 900K
C. 600K
D. 150K
Answer
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Hint: Root mean square velocity or r.m.s velocity is the square root of the mean of the squares of the several velocities of the gas molecules. It is indicated 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 of the gas
M = Gram molecular mass of the gas
Complete Step by Step Solution:
The gas molecules remain by broad distances and are independent of motion in the area in which it is encompassed.
These molecules have constant motion in different directions at varied velocities.
Collisions occur between these molecules and the walls of the container which is defined as a molecular collision.
The expression for r.m.s velocity is:-
\[{{\rm{V}}_{{\rm{rms}}}}{\rm{ = }}\sqrt {\frac{{{\rm{3RT}}}}{{\rm{M}}}} \]
Given
\[{{\rm{V}}_{{\rm{rms}}}}\]=v
Temperature=300K
We have to find out the temperature at which the root mean square velocity becomes double.
From the expression for r.m.s velocity, we see that r.m.s velocity and the square root of temperature have a direct relationship.
For the given gas, the relationship between the root mean square velocity at 300K and T is given by the expression:-
\[\frac{{{{\rm{V}}_{\rm{1}}}}}{{{{\rm{V}}_{\rm{2}}}}}{\rm{ = }}\sqrt {\frac{{{{\rm{T}}_{\rm{1}}}}}{{{{\rm{T}}_{\rm{2}}}}}} \]
where
\[{{\rm{V}}_{\rm{1}}}\]=r.m.s velocity at 300K=v
\[{{\rm{V}}_{\rm{2}}}\]=r.m.s velocity at a temperature which we need to find out=2v
T1=300K
We have to find out T2.
Squaring both sides we get,
\[{\left( {\frac{{\rm{v}}}{{{\rm{2v}}}}} \right)^{\rm{2}}}{\rm{ = }}\frac{{{{\rm{T}}_{\rm{1}}}}}{{{{\rm{T}}_{\rm{2}}}}}\]
Putting the given values in the above equation
\[{\left( {\frac{{\rm{1}}}{{\rm{2}}}} \right)^{\rm{2}}}{\rm{ = }}\frac{{{\rm{300}}}}{{{{\rm{T}}_{\rm{2}}}}}\]
\[ \Rightarrow {{\rm{T}}_{\rm{2}}}{\rm{ = 300 \times 4 = 1200K}}\]
Therefore, the temperature of the gas at which the r.m.s velocity is doubled is 1200K.
So, option A is correct.
Additional Information: Root mean square velocity is expressed in m/s. Temperature and molecular mass are taken in kelvin and kilograms respectively.
Note: With the increase in the square root of molecular mass, root mean square velocity decreases. So, gases with low molecular mass have a high root mean square velocity. Hence, they move faster as compared to gases of high molecular mass. There is no effect of change in pressure or volume on the rms speed as at a given temperature, PV is constant.
Formula Used:
\[{{\rm{V}}_{{\rm{rms}}}}{\rm{ = }}\sqrt {\frac{{{\rm{3RT}}}}{{\rm{M}}}} \]
where
R = Universal gas constant
T = Temperature of the gas
M = Gram molecular mass of the gas
Complete Step by Step Solution:
The gas molecules remain by broad distances and are independent of motion in the area in which it is encompassed.
These molecules have constant motion in different directions at varied velocities.
Collisions occur between these molecules and the walls of the container which is defined as a molecular collision.
The expression for r.m.s velocity is:-
\[{{\rm{V}}_{{\rm{rms}}}}{\rm{ = }}\sqrt {\frac{{{\rm{3RT}}}}{{\rm{M}}}} \]
Given
\[{{\rm{V}}_{{\rm{rms}}}}\]=v
Temperature=300K
We have to find out the temperature at which the root mean square velocity becomes double.
From the expression for r.m.s velocity, we see that r.m.s velocity and the square root of temperature have a direct relationship.
For the given gas, the relationship between the root mean square velocity at 300K and T is given by the expression:-
\[\frac{{{{\rm{V}}_{\rm{1}}}}}{{{{\rm{V}}_{\rm{2}}}}}{\rm{ = }}\sqrt {\frac{{{{\rm{T}}_{\rm{1}}}}}{{{{\rm{T}}_{\rm{2}}}}}} \]
where
\[{{\rm{V}}_{\rm{1}}}\]=r.m.s velocity at 300K=v
\[{{\rm{V}}_{\rm{2}}}\]=r.m.s velocity at a temperature which we need to find out=2v
T1=300K
We have to find out T2.
Squaring both sides we get,
\[{\left( {\frac{{\rm{v}}}{{{\rm{2v}}}}} \right)^{\rm{2}}}{\rm{ = }}\frac{{{{\rm{T}}_{\rm{1}}}}}{{{{\rm{T}}_{\rm{2}}}}}\]
Putting the given values in the above equation
\[{\left( {\frac{{\rm{1}}}{{\rm{2}}}} \right)^{\rm{2}}}{\rm{ = }}\frac{{{\rm{300}}}}{{{{\rm{T}}_{\rm{2}}}}}\]
\[ \Rightarrow {{\rm{T}}_{\rm{2}}}{\rm{ = 300 \times 4 = 1200K}}\]
Therefore, the temperature of the gas at which the r.m.s velocity is doubled is 1200K.
So, option A is correct.
Additional Information: Root mean square velocity is expressed in m/s. Temperature and molecular mass are taken in kelvin and kilograms respectively.
Note: With the increase in the square root of molecular mass, root mean square velocity decreases. So, gases with low molecular mass have a high root mean square velocity. Hence, they move faster as compared to gases of high molecular mass. There is no effect of change in pressure or volume on the rms speed as at a given temperature, PV is constant.
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