Question

The temperature of a gas is doubled (i) on absolute scale (ii) on centigrade scale. The increase in root mean square velocity of gas will be:
A. more in case (i)
B. more in case (ii)
C. same in both cases
D. information not sufficient.

Answer 1

Hint: Root mean square of velocity is the square root of average velocity. It is used to find how fast molecules are moving at a given temperature. How fast molecules move is directly proportional to their absolute temperature and inversely proportional to their mass.

Complete step by step answer:
If ${{\text{v}}_1}{\text{,}}{{\text{v}}_2}{\text{,}}{{\text{v}}_3}{\text{,}}...{{\text{v}}_{\text{n}}}$ are the velocities of ${\text{n}}$ molecules in a gas, ${\mu ^2}$ be the mean squares of all the velocities, then
${\mu ^2} = \dfrac{{{{\text{v}}_1}^2 + {{\text{v}}_2}^2 + {{\text{v}}_3}^2 + ... + {{\text{v}}_{\text{n}}}^2}}{{\text{n}}}$
Taking the root, we get the root mean square velocity.
Root mean square velocity, $\mu = \sqrt {\dfrac{{{{\text{v}}_1}^2 + {{\text{v}}_2}^2 + {{\text{v}}_3}^2 + ... + {{\text{v}}_{\text{n}}}^2}}{{\text{n}}}} $
The value of the RMS velocity $\left( \mu \right)$, at a given temperature can be calculated from the kinetic gas equation:
${\text{PV}} = \dfrac{1}{3}{\text{mN}}{{\text{u}}^2}$, where ${\text{P}}$ is the pressure, ${\text{V}}$ is the volume of gas, ${\text{m}}$ is the mass, ${\text{N}}$ is the number of molecules, ${\text{u}}$ is the velocity.
From this equation, ${{\text{u}}^2} = \dfrac{{3{\text{PV}}}}{{{\text{mN}}}} \to \left( 1 \right)$
For one mole of gas, ${\text{n}} = 1$, ideal gas equation can be written as:
${\text{PV}} = {\text{RT}} \to \left( 2 \right)$
Substitute $\left( 2 \right)$ in $\left( 1 \right)$, we get
${{\text{u}}^2} = \dfrac{{3{\text{RT}}}}{{\text{M}}}$, where ${\text{M}}$ is the molar mass since the product of number of molecules and mass is the molar mass.
Thus taking root, ${\text{u}} = \sqrt {\dfrac{{3{\text{RT}}}}{{\text{M}}}} $
(i) Here, the temperature doubles which indicates that let the temperature changes from ${{\text{T}}_1}{\text{ = 100K}}$ to ${{\text{T}}_2} = 200{\text{K}}$.
Increase in rms velocity will be expressed as the ratio of final rms velocity to initial rms velocity.
i.e. $\dfrac{{{{\text{u}}_2}}}{{{{\text{u}}_1}}} = \sqrt {\dfrac{{3{\text{R}}{{\text{T}}_1}}}{{\text{M}}}} \div \sqrt {\dfrac{{3{\text{R}}{{\text{T}}_2}}}{{\text{M}}}} = \dfrac{{{{\text{T}}_2}}}{{{{\text{T}}_1}}}$
Substituting the values of temperature, we get
\[\dfrac{{{{\text{u}}_2}}}{{{{\text{u}}_1}}} = \sqrt {\dfrac{{200{\text{K}}}}{{{\text{100K}}}}} = \sqrt 2 = 1.414\]
(ii) When temperature is in centigrade, let the temperature changes from ${{\text{T}}_1} = {100^ \circ }{\text{C = 273K + 100 = 373K}}$ to ${{\text{T}}_2} = {200^ \circ }{\text{C = 273K + 200 = 473K}}$.
Increase in rms velocity can be expressed as:
\[\dfrac{{{{\text{u}}_2}}}{{{{\text{u}}_1}}} = \sqrt {\dfrac{{{\text{473K}}}}{{{\text{373K}}}}} = \sqrt {1.27} = 1.13\]
Comparing both values, value in absolute scale is more than that in centigrade.
Thus, the correct option is A.

Note:
There are three kinds of velocities. They are average velocity, root mean square velocity and most probable velocity.
Root mean square is a more accurate way of defining the average speed. At the same temperature, small molecules move fast and large molecules move slow.