Kinetic Theory of Gases MCQ Questions & Answers in Heat and Thermodynamics | Physics
Learn Kinetic Theory of Gases MCQ questions & answers in Heat and Thermodynamics are available for students perparing for IIT-JEE, NEET, Engineering and Medical Enternace exam.
11.
The temperature of an ideal gas is increased from $$120\,K$$ to $$480\,K.$$ If at $$120\,K$$ the root-mean-square velocity of the gas molecules is $$v,$$ at $$480\,K$$ it becomes
12.
A polyatomic gas with $$n$$ degrees of freedom has a mean energy per molecule given by
A
$$\frac{{nkT}}{N}$$
B
$$\frac{{nkT}}{2N}$$
C
$$\frac{{nkT}}{2}$$
D
$$\frac{{3kT}}{2}$$
Answer :
$$\frac{{nkT}}{2}$$
According to law of equipartition of energy, the energy per degree of freedom is $$\frac{1}{2}kT.$$ For a polyatomic gas with $$n$$ degrees of freedom, the mean energy per molecule $$ = \frac{1}{2}nkT$$
13.
Boyle' law is applicable for an
A
adiabatic process.
B
isothermal process.
C
isobaric process.
D
isochoric process
Answer :
isothermal process.
According to boyle's law, at constant temperature.
$$P \propto \frac{1}{{\;V}}\,\,{\text{or}}\,\,{P_1}\;{V_1} = {P_2}\;{V_2}$$
14.
A non-linear triatomic gas is filled inside a vessel. If $$'\alpha '$$ fraction of moles dissociate into individual atoms, then average degree of freedom for the mixture is : (neglect vibrational degrees of freedom)
15.
The gases carbon-monoxide $$\left( {CO} \right)$$ and nitrogen at the same temperature have kinetic energies $${E_1}$$ and $${E_2}$$ respectively. Then
A
$${E_1} = {E_2}$$
B
$${E_1} > {E_2}$$
C
$${E_1} < {E_2}$$
D
$${E_1}$$ and $${E_2}$$ cannot be compared
Answer :
$${E_1} = {E_2}$$
The gases carbon-monoxide $$\left( {CO} \right)$$ and nitrogen $$\left( {{N_2}} \right)$$ are diatomic, so both have equal kinetic energy $$\frac{5}{2}kT,{\text{i}}{\text{.e}}{\text{.}}\,{E_1} = {E_2}.$$
16.
The value of $${C_p} - {C_v}$$ is $$1.00R$$ for a gas sample in state $$A$$ and is $$1.06R$$ in state $$B.$$ Let $${p_A},{p_B}$$ denote the pressure and $${T_A},{T_B}$$ denote the temperature of the states $$A$$ and $$B$$ respectively. Then most likely
For ideal gas $${C_p} - {C_v} = R$$
If $${C_p} - {C_v} = 1.06\,R,$$
then gas will be real gas. Thus pressure is high and temperature is low for real gas.
17.
The pressure of a gas is raised from $${27^ \circ }C$$ to $${927^ \circ }C.$$ The root mean square speed
A
is $$\sqrt {\left( {\frac{{927}}{{27}}} \right)} $$ times the earlier value
B
remains the same
C
gets halved
D
gets doubled
Answer :
gets doubled
$$RMS$$ speed is defined as the square root of the mean of the squares of the random velocities of the individual molecules of a gas. From Maxwellian distribution law, $$RMS$$ speed is given by $${c_{rms}} = \sqrt {\left( {\frac{{3kT}}{m}} \right)} $$
$$ \Rightarrow {c_{rms}} \propto \sqrt T $$
For two different cases i.e. at two different temperatures
$$\therefore \frac{{{{\left( {{c_{rms}}} \right)}_1}}}{{{{\left( {{c_{rms}}} \right)}_2}}} = \sqrt {\frac{{{T_1}}}{{{T_2}}}} $$
Here, $${T_1} = {27^ \circ }C = 300\,K$$
$$\eqalign{
& {T_2} = {927^ \circ }C = 1200\,K \cr
& \therefore \frac{{{{\left( {{c_{rms}}} \right)}_1}}}{{{{\left( {{c_{rms}}} \right)}_2}}} = \sqrt {\frac{{300}}{{1200}}} = \frac{1}{2} \cr
& \Rightarrow {\left( {{c_{rms}}} \right)_2} = 2{\left( {{c_{rms}}} \right)_1} \cr} $$
Hence, root mean square speed will be doubled.
18.
The equation of state for $$5g$$ of oxygen at a pressure $$p$$ and temperature $$T,$$ when occupying a volume $$V,$$ will be
Number of moles, $$n = \frac{m}{{{\text{molecular weight}}}} = \frac{5}{{32}}$$
As, from ideal gas equation $$pV = nRT \Rightarrow pV = \frac{5}{{32}}RT$$
19.
The mean free path of molecules of a gas, (radius $$'r'$$) is inversely proportional to :
A
$${r^3}$$
B
$${r^2}$$
C
$$r$$
D
$$\sqrt r $$
Answer :
$${r^2}$$
Mean free path $${\lambda _m} = \frac{1}{{\sqrt 2 \pi {d^2}n}}$$
where $$d$$ = diameter of molecule and $$d = 2r$$
$$\therefore {\lambda _m} \propto \frac{1}{{{r^2}}}$$
20.
The average translational energy and the $$rms$$ speed of molecules in a sample of oxygen gas at $$300\,K$$ are $$6.21 \times {10^{ - 21}}J$$ and $$484\,m/s$$ respectively. The corresponding values at $$600\,K$$ are nearly (assuming ideal gas behavior)
A
$$12.42 \times {10^{ - 21}}J,968\,m/s$$
B
$$8.78 \times {10^{ - 21}}J,684\,m/s$$
C
$$6.21 \times {10^{ - 21}}J,968\,m/s$$
D
$$12.42 \times {10^{ - 21}}J,684\,m/s$$
Answer :
$$12.42 \times {10^{ - 21}}J,684\,m/s$$
The formula for average kinetic energy is
$$\eqalign{
& \overline {K.E} = \frac{3}{2}kT \cr
& \therefore \frac{{{{\left( {\overline {K.E} } \right)}_{600\;K}}}}{{{{\left( {\overline {K.E} } \right)}_{300\;K}}}} = \frac{{600}}{{300}} \cr
& \Rightarrow {\left( {\overline {K.E} } \right)_{600\;K}} = 2 \times 6.21 \times {10^{ - 21}}\,J \cr
& = 12.42 \times {10^{ - 21}}\,J \cr} $$
Also the formula for $$r.m.s.$$ velocity is
$$\eqalign{
& {C_{rms}} = \sqrt {\frac{{3KT}}{m}} \cr
& \therefore \frac{{{{\left( {{C_{ms}}} \right)}_{600\;K}}}}{{{{\left( {{C_{rms}}} \right)}_{300\;K}}}} = \sqrt {\frac{{600}}{{300}}} \cr
& \Rightarrow {\left( {{C_{rms}}} \right)_{600\;K}} = \sqrt 2 \times 484 = 684\,m/s \cr} $$