Kinetic Theory of Gases MCQ Questions & Answers in Heat and Thermodynamics | Physics

$$\eqalign{ & \frac{{{V_N}}}{{{V_{He}}}} = \sqrt {\frac{{{\gamma _{{N_2}}}{M_{He}}}}{{{\gamma _{He}}{M_{{N_2}}}}}} \cr & = \sqrt {\frac{{\frac{7}{5} \times 4}}{{\frac{5}{3} \times 28}}} \cr & = \frac{{\sqrt 3 }}{5} \cr} $$

$$\eqalign{ & {v_{rms}} = {\left[ {\frac{{{{\left( 2 \right)}^2} + {{\left( 3 \right)}^2} + {{\left( 4 \right)}^2} + {{\left( 5 \right)}^2}}}{4}} \right]^{\frac{1}{2}}} \cr & = \sqrt {\left[ {\frac{{54}}{4}} \right]} \cr} $$

Note : At constant volume, Charle's law is used.

$$\eqalign{ & {P_1}{M_1} = {d_1}RT;\,{P_2}{M_2} = {d_2}RT \cr & \therefore \,\,\frac{{{P_1}}}{{{P_2}}} \times \frac{{{M_1}}}{{{M_2}}} = \frac{{{d_1}}}{{{d_2}}} \cr & \frac{4}{3} \times \frac{2}{3} = \frac{{{d_1}}}{{{d_2}}} \cr & \therefore \,\,\frac{{{d_1}}}{{{d_2}}} = \frac{8}{9} \cr} $$

Mean translational $$K.E.$$  of any molecule whether monoatomic, diatomic or polyatomic, at a given temperature $$T$$ is $$\frac{3}{2}kT.$$

In elastic collision kinetic energy is conserved.

$${v_{rms}} = \sqrt {\frac{{8RT}}{{\pi M}}} $$
For $${v_{rms}}$$ to be equal $$\frac{{{T_{{H_2}}}}}{{{M_{{H_2}}}}} = \frac{{{T_{{O_2}}}}}{{{M_{{O_2}}}}}$$
Here, $${M_{{H_2}}} = 2;{M_{{O_2}}} = 32;$$
$$\eqalign{ & {T_{{O_2}}} = 47 + 273 = 320\,K \cr & \therefore \,\,\frac{{{T_{{H_2}}}}}{2} = \frac{{320}}{{32}} \cr & \Rightarrow \,\,{T_{{H_2}}} = 20\,K \cr} $$

If $$T$$ be the required temperature, then,
$$\eqalign{ & {v_{rms}} - {v_p} = 200\,m/s \cr & {\text{or}}\,\,\sqrt {\frac{{3RT}}{{{M_{He}}}}} - \sqrt {\frac{{2RT}}{{{M_{He}}}}} = 200\,m/s \cr & {\text{or}}\,\,\sqrt T \left[ {\sqrt 3 - \sqrt 2 } \right] = \sqrt {\frac{{{M_{He}}}}{R}} \left( {200\,m/s} \right) \cr} $$
Substituting $${M_{He}} = 4 \times {10^{ - 3}}\,kg,R = 8.314\,J/mol - K$$
$$T = 190.2\,K.$$

The state of a homogeneous system at any time is described in terms of three thermodynamic parameters viz pressure $$\left( p \right),$$ volume $$\left( V \right)$$ and temperature $$\left( T \right).$$ The mathematical relation between these parameters is called the equation of state of the thermodynamic system.
For $$n$$ moles of an ideal gas, the equation of state of the thermodynamic system is
$$pV = nRT\,......\left( {\text{i}} \right)$$
where $$R$$ is universal gas constant for one gram mole of an ideal gas.
Eq. (i) can also be expressed as
$$\eqalign{ & pV = nRT \cr & n = \frac{{{\text{wt}}{\text{.}}\,{\text{of}}\,{O_2}}}{{{\text{molecular weight of }}{O_2}}} = \frac{8}{{32}} = \frac{1}{4} \cr & {\text{So,}}\,\,pV = \frac{1}{4}RT \cr} $$

$${c_{rms}} \propto \sqrt T $$
As temperature increases from $$300\,K$$  to $$1200\,K$$  that is four times, so, $${c_{rms}}$$ Will be doubled.