MCQ Questions for Class 11 Engineering Physics Kinetic Theory Quiz 5

A uniform tube with piston in the middle and containing a gas at 0 $^{0}$ C is heated to 100 $^{0}$ C at one side . If the piston moves 5 cm, find the length of the tube containing the gas at 100 $^{0}$ C.

0%
16.75 cm
0%
37.5 cm
0%
28.25 cm
0%
12.5 cm

Explanation

$\dfrac{PV}{T}$ is constant for gas in both sections of the cylinder.

Let the length and area of the tube be

$L$ and

$A$ , respectively.

For section with constant temperature,

$P_{0}V_{0}=P_1V_1$

$\implies P_ (A\dfrac{L}{2})=P_1(A(\dfrac{L}{2}-5))$

Initial temperature

$T_0 = 0^oC = 273 \textrm{K}$

Final temperature

$T_1 = 100^oC = 100+273 = 373 \textrm{K}$

Similarly for section whose temperature increased,

$\dfrac{P_0V_0}{T_0}=\dfrac{P_1V_1}{T_1}$

$\implies \dfrac{P_0(A\dfrac{L}{2})}{273}=\dfrac{P_1(A(\dfrac{L}{2}+5))}{373}$

$\implies L=65cm$

So the length of tube containing gas at

$100^{\circ}$ is

$\dfrac{L}{2}+5=37.5cm$

In a given species of tobbaco there is 0.1 mg of virus per c.c. The mass of virus is $4 \times 10^{7}$ Kg per kilomol. The number of the molecules of virus present in 1 c.c. will be

0%
$10^{13}$
0%
$10^{11}$
0%
$1.5 \times 10^{12}$
0%
$10^{14}$

Explanation

${ m }_{ v }=\dfrac { M }{ { N }_{ A } } =\dfrac { 4\times { 10 }^{ 7 } }{ 6\times { 10 }^{ 23 }\times { 10 }^{ 3 } } \\ mass\quad of\quad virus\quad per\quad cc=0.1\times { 10 }^{ -6 }kg\\$

Number of molecules in cc

$n=\dfrac { { m }_{ v } }{ { N }_{ A } } =\dfrac { 0.1\times { 10 }^{ -6 }\times 6\times { 10 }^{ 23 }\times { 10 }^{ 3 } }{ 4\times { 10 }^{ 7 } } \\ n=1.5\times { 10 }^{ 12 }$

A vertical cylindrical vessel separated in two parts by a frictionless piston free to move along the length of a vessel. The length of vessel is 90 cm and the piston divides the cylinder in the ratio 2 :Each of the two parts contain 0.1 mole of an ideal gas. The temperature of the gas is $27^{o}$ C in each part. The mass of the piston is

0%
41.5 kg
0%
8.97 kg
0%
9.57 kg
0%
11.57 kg

Explanation

$PV=nRT$

$P = nRT/V$

$PA=nRT/l$

$PA=0.1\times 8.31\times 300/0.6$

$PA=41.5 N$
Therefore option(A) is correct.

At what temperature the linear kinetic energy of a gas molecule will be equal to that of an electron accelerated through a potential difference of 10 volt?

0%
77.3 x 10 $^{3}$ K
0%
38.65 x 10 $^{3}$ K
0%
19.x10 $^{3}$ K
0%
273 K

Explanation

Energy of an electron accelerated through a potential V:

$E=eV$

Given,

$E=\dfrac{3}{2}kT$

$\Rightarrow eV = \dfrac{3}{2}kT$

$\Rightarrow T =\dfrac{2eV}{3k} =\dfrac{2e \times 10}{3 \times 1.38 \times 10^{-23}}= \dfrac{20\: eV}{3 \times 1.38 \times 10^{-23}}=\dfrac{20 \times 1.6 \times 10^{-19}J}{3 \times 1.38 \times 10^{-23}}=77.3 \times 10^3\:K$

One gram mol of helium at 27

$^{0}$ C is mixed with three gram mols of oxygen at 127

$^{0}$ C at constant pressure. If there is no exchange of heat with the atmosphere then the final temperature will be

0%
375 K
0%
175K
0%
475 K
0%
575K

The mean molecular energy of gas at 300 K will be

0%
6.2 x 10 $^{20}$ Joule
0%
6.2 x 10 $^{-20}$ Joule
0%
6.2 x 10 $^{-21}$ Joule
0%
2.6 x 10 $^{-20}$ Joule

Explanation

Average kinetic energy:

$(KE)_{avg}= \frac{3}{2}kT=\frac{3}{2}\times 1.38 \times 10^{-23} \times 300=6.2 \times 10^{-21}J$ where k is
Boltzman's constant and

$T=(30 + 273)^{\circ}K= 300^{\circ}K$

The degrees of freedom of a diatomic gas at normal temperature is

0%
3
0%
4
0%
5
0%
6

Explanation

In three-dimensional space, three degrees of freedom are associated with the movement of a particle. A diatomic gas molecule thus has 6 degrees of freedom. This set may be decomposed in terms of translations, rotations, and vibrations of the molecule. The center of mass motion of the entire molecule accounts for 3 degrees of freedom. In addition, the molecule has two rotational degrees of motion and one vibrational mode The rotations occur around the two axes perpendicular to the line between the two atoms. The rotation around the atom-atom bond is not a physical rotation. At normal temp, vibration is not possible. Hence, the total no of degrees of freedom is

$f= 3+ 2=5$

Equal masses of $N_{2}$ and $O_{2}$ gases are filled in vessel A and B. The volume of vessel B is double of A. The ratio of pressure in vessel A and B will be

0%
16:7
0%
16:14
0%
32:7
0%
32:28

Explanation

The mathematical form of the Ideal Gas Law is:
PV = nRT and n = m/MW
Where:P - pressure
V - volume
n - number of moles
T - temperature
m - mass
MW - Molecular Weight
R - ideal gas constant
So we can write,

$\dfrac{P_a}{ P_b}$ =

$\dfrac{M_b \times V_b}{ M_a \times V_a}$
here,

$V_b = 2 V_a$ and

$M_a$ = atomic wt of

$O_2$ = 8 and

$M_b$ = atomic wt of

$N_2$ = 7
Putting these values we get,

$P_a/ P_b$ = 16/7

A partially inflated balloon contains 500 m $^{3}$ of helium at 27 $^{0}$ C and 1 atm pressure. The volume of the helium at an altitude of 6000 m, where the pressure is 0.5atm and the temperature is -3 $^{0}$ C is

0%
22.4 lit
0%
11.2 lit
0%
900 m $^{3}$
0%
50 m $^{3}$

Explanation

Assuming helium to be an ideal gas, and using relation:

$\dfrac{P_1V_1}{T_1}=\dfrac{P_2V_2}{T_2}$

$V_2=V_1\dfrac{P_1T_2}{P_2T_1}=900m^3$

The temperature of gas is increased from 27

$^{o}$ C to 127

$^{o}$ C. The ratio of its mean kinetic energies will be

0%
$\dfrac{3}{4}$
0%
$\dfrac{4}{3}$
0%
$\dfrac{9}{16}$
0%
$\dfrac{16}{9}$

Explanation

We know that , K.E

$\alpha$ T ; where, T is absolute temperature
So, ratio of kinetic energies = ratio of temperatures (in Kelvin)
=

$\dfrac{300K}{400K} = \dfrac{3}{4}$

Figure shows a cylindrical tube of cross-sectional area A filled with two frictionless pistons. The pistons are connected through wire. The tension in the wire if the temperature rises from T $_{0}$ to 2T $_{0}$ is (Initial pressure is P $_{0}$ , atmospheric pressure)

0%
$P_{0}A$
0%
$\frac{P_{0}A}{2}$
0%
$2P_{0}A$
0%
$\frac{2P_{0}A}{3}$

Explanation

Since the wired would be taut, there would be no change in the volume of enclosed gas, so

$\dfrac{P}{T}=constant$

$\implies \dfrac{P}{2T_0}=\dfrac{P_0}{T_0}$

$\implies P=2P_0$

Thus, an equilibrium to exist, the difference in inside and outside forces must be balanced by the tension in the wire.

Thus,

$T=2P_0A-P_0A=P_0A$

28gm of $N_2$ gas is contained in a flask at a pressure of 10atm and at a temperature of 57 $^{0}$ C. It is found that due to leakage in the flask, the pressure is reduced to half and the temperature reduced to 27 $^{0}$ C. The quantity of N $_{2}$ gas that leaked out is

0%
$\frac{5}{63}gm$
0%
$\frac{63}{5}gm$
0%
$\frac{11}{20}gm$
0%
$\frac{20}{11}gm$

Explanation

Since the volume of the flask remains the same, using Ideal Gas Equation we claim,

$n\propto \dfrac{P}{T}$

$\implies \dfrac{n_2}{n_1}=\dfrac{P_2T_1}{P_1T_2}$

$\implies n_2=\dfrac{11}{20}n_1$

Therefore, fraction of gas leaked out is

$(1-\dfrac{11}{20})=\dfrac{9}{20}$

Thus, amount of gas leaked out=

$\dfrac{9}{20}\times 28gm=\dfrac{63}{5}gm$

$Assertion$ : gases are characterised with two coefficients of expansion

$Reason$ : when heated both volume and pressure increase with the rise in temperature

0%
Both assertion (A) and reason (R) are correct and R gives the correct explanation
0%
Both assertion (A) and reason (R) are correct but R doesnt give the correct explanation
0%
A is true but R is false
0%
A is false but R is true

Explanation

The ideal gas equation

$PV=\eta RT$ tells us that the pressure and volume are directly proportional to the temperature. The coefficients of expansion for pressure is

$\gamma_p$ and that of volume is

$\alpha_v$ .

Assertion: Gasses obey Boyle's law at high temperature and low pressure only.

Reason: At low pressure and high temperature, gasses would behave like ideal gases.

0%
Both assertion (A) and reason (R) are correct and R gives the correct explanation
0%
Both assertion (A) and reason (R) are correct but R does give the correct explanation
0%
A is true but R is false
0%
A is false but R is true

Follwing operation are carried out on a sample of ideal gas initially at pressure P volume V and kelvin temperature T.

a) At constant volume, the pressure is increased fourfold.

b) At constant pressure, the volume is doubled

c) The volume is doubled and pressure halved.

d) If heated in a vessel open to atmosphere, one-fourth of the gas escapes from the vessel.

Arrange the above operations in the increasing order of final temperature.

0%
a,b,c,d
0%
c,b,a,d
0%
b,a,d,c
0%
c,d,b,a

Explanation

ideal gas equation

$PV = nRT$ eq(1)

where

$P$ = Pressure

$V$ = Volume

$n$ = number of moles

$R$ = universal gas constant

$T$ = Temperature

suppose initial values are

${P}_{1},{V}_{1},{n}_{1},{T}_{1}$ and

final values are

${P}_{2},{V}_{2},{n}_{2},{T}_{2}$

${P}_{1}{V}_{1}={n}_{1}R{T}_{1}$ eq(2)

${P}_{2}{V}_{2}={n}_{2}R{T}_{2}$ eq(3)

(a) At constant volume, the pressure is increased fourfold

${V}_{2} = {V}_{1}$

${P}_{2} = 4 {P}_{1}$

${n}_{2} = {n}_{1}$

$\dfrac{eq(3)}{eq(2)}$

$\dfrac{{T}_{2}}{{T}_{1}} = 4$

${T}_{2} = 4 {T}_{1}$

(b) At constant pressure, the volume is doubled

${V}_{2} = 2 {V}_{1}$

${P}_{2} = {P}_{1}$

${n}_{2} = {n}_{1}$

$\dfrac{eq(3)}{eq(2)}$

$\dfrac{{T}_{2}}{{T}_{1}} = 2$

${T}_{2} = 2 {T}_{1}$

${V}_{2} = 2 {V}_{1}$

${P}_{2} =\dfrac{1}{2} {P}_{1}$

${n}_{2} = {n}_{1}$

$\dfrac{eq(3)}{eq(2)}$

$\dfrac{{T}_{2}}{{T}_{1}} = 1$

${T}_{2} = {T}_{1}$

(d) If heated in a vessel open to atmosphere, onefourth of the gas escapes from the vessel.

for open vessel pressure and volume will be constant.

${V}_{2} = {V}_{1}$

${P}_{2} = {P}_{1}$

one-fourth of gas escaped from vessel, so three-fourth gas left

${n}_{2} = \dfrac{3}{4}{n}_{1}$

$\dfrac{\dfrac{3}{4} {T}_{2}}{{T}_{1}} = 1$

${T}_{2} = \dfrac{4}{3}{T}_{1}$

arranging final temperature in increasing order

(c),(d),(b),(a)

PV = n RT holds good for :

a) Isobaric process b) Isochoric process

c) Isothermal process d) Adiabatic process

0%
a & b
0%
a,b & c
0%
a,b & d
0%
all

Assertion: In Joules bulb apparatus, as reservoir is moved up, the mercury level raises into the bulb.

Reason: The pressure on the enclosed gas increases.

0%
Both assertion (A) and reason (R) are correct and R gives the correct explanation
0%
Both assertion (A) and reason (R) are correct but R doesnt give the correct explanation
0%
A is true but R is false
0%
A is false but R is true

Assertion:With increase in temperature, the pressure of given gas increases

Reason:Increase in temperature causes decrease in no. of collision of molecules with walls of container.

0%
Both assertion (A) and reason (R) are correct and R gives the correct explanation
0%
Both assertion (A) and reason (R) are correct but R doesnt give the correct explanation
0%
A is true but R is false
0%
A is false but R is true

According to kinetic theory of gasses at Zero kelvin

a) Pressure of ideal gas is zero

b) Volume of ideal gas is zero

c) Internal energy of ideal gas is zero

d) Matter exists in gaseous state only

0%
a & d are true
0%
a,b & d are true
0%
a,b & c are true
0%
All are true

Which of the following processes will quadruple the pressure

a) Reduce V to half and double T

b) Reduce V to 1/8th and reduce T to half

c) Double V and half T

d) Increase both V and T to double the values

0%
b,c
0%
a,b
0%
c,d
0%
a,d.

Assertion:PV/T=constant for 1 mole of gas. This constant is same for all gases.

Reason:1 mole of different gases at NTP occupy same volume of 22.4 litres.

0%
Both assertion (A) and reason (R) are correct and R gives the correct explanation
0%
Both assertion (A) and reason (R) are correct but R doesnt give the correct explanation
0%
A is true but R is false
0%
A is false but R is true

Assertion: PV/T=constant for 1 gram of gas. This constant varies from gas to gas.

Reason:1 gram of different gases at NTP occupy different volumes.

0%
Both assertion (A) and reason (R) are correct and R gives the correct explanation
0%
Both assertion (A) and reason (R) are correct but R doesnt give the correct explanation
0%
A is true but R is false
0%
A is false but R is true

To find out degree of freedom, the correct expression is :

0%
$f=\dfrac { 2 }{ \gamma -1 }$
0%
$f=\dfrac { \gamma +1 }{ 2 }$
0%
$f=\dfrac { 2 }{ \gamma +1 }$
0%
$f=\dfrac { 1 }{ \gamma +1 }$

Explanation

$\because \gamma =1+\dfrac { 2 }{ f }$

$\Longrightarrow \dfrac { 2 }{ f } =\gamma -1\Longrightarrow f=\dfrac { 2 }{ \gamma -1 }$

A vessel contains air and saturated vapor. The pressure of air is

$\mathrm{p}_{2}$ and

$\mathrm{p}_{1}$ is the S.V. P. On compressing the mixture to one-fourth of its original volume, what is the increase in pressure of the mixture?

0%
$2\mathrm{p}_{1}$
0%
$2\mathrm{p}_{2}$
0%
$3\mathrm{p}_{1}$
0%
$3\mathrm{p}_{2}$

Explanation

Initially total pressure

$=(P_{1}+P_{2})$
When compressed to

$\dfrac{1}{4}$ of its volume, the pressure of air becomes

$4P_{2}.$
Pressure of saturated vapour remains the same.
Total pressure after compression

$=4P_{2}+P_{1}$
Difference in pressure

$=(4P_{2}+P_{1})-(P_{1}+P_{2})$

$=3P_{2}$

Match List I with List II

List-I List-II

a) 0.00366/ $^{0}$ C e) Avagadros Number

b) 6.023x10 $^{23}$ molecules f) Universal gas constant

c) -273 $^{0}$ C g) Pressure coefficient of gas

d) 8.31 J/K-mole h) Intercept of V-T graph at

constnat pressure

0%
a-g, b-e, c-h, d-f.
0%
a-f, b-g c-e,d-h.
0%
a-g,b-e,c-f, d-h.
0%
a-g,b-f, c-e, d-h.

Explanation

(a)

$0.00366^{o}C$ =

$\frac{1}{273}^{o}C$ = pressure coefficient of gas
a - g
(b)

$6.023 \times {10}^{23}$ is Avagadros Number.
b - e
(c)

$-273^{o}C$ is Intercept of V-T graph at constnat pressure(as shown in image)
c - h
(d)8.31 J/K-mole is Universal gas constant(R).
d - f
Answer is A.

Three perfect gases at absolute temperatures

$\mathrm{T}_{1},\ \mathrm{T}_{2}$ and

$\mathrm{T}_{3}$ are mixed. The masses of molecules are

$\mathrm{m}_{1},\ \mathrm{m}_{2}$ and

$\mathrm{m}_{3}$ and the number of molecules are

$\mathrm{n}_{1},\ \mathrm{n}_{2}$ and

$\mathrm{n}_{3}$ respectively. Assuming no loss of energy, the final temperature of the mixture is:

0%
$\displaystyle \frac{(\mathrm{T}_{1}+\mathrm{T}_{2}+\mathrm{T}_{3})}{3}$
0%
$\displaystyle \frac{\mathrm{n}_{1}\mathrm{T}_{\mathrm{l}}+\mathrm{n}_{2}\mathrm{T}_{2}+\mathrm{n}_{3}\mathrm{T}_{3}}{\mathrm{n}_{1}+\mathrm{n}_{2}+\mathrm{n}_{3}}$
0%
$\displaystyle \frac{\mathrm{n}_{1}\mathrm{T}_{1}^{2}+\mathrm{n}_{2}\mathrm{T}_{2}^{2}+\mathrm{n}_{3}\mathrm{T}_{3}^{3}}{\mathrm{n}_{1}\mathrm{T}_{1}+\mathrm{n}_{2}\mathrm{T}_{2}+\mathrm{n}_{3}\mathrm{T}_{3}}$
0%
$\displaystyle \frac{\mathrm{n}_{1}^{2}\mathrm{T}_{1}^{2}+\mathrm{n}_{2}^{2}\mathrm{T}_{2}^{2}+\mathrm{n}_{3}\mathrm{T}_{3}^{3}}{\mathrm{n}_{1}\mathrm{T}_{1}+\mathrm{n}_{2}\mathrm{T}_{2}+\mathrm{n}_{3}\mathrm{T}_{3}}$

Explanation

Number of moles of first gas

$=\displaystyle \frac{\mathrm{n}_{1}}{\mathrm{N}_{\mathrm{A}}}$
Number of moles of second gas

$=\displaystyle \frac{\mathrm{n}_{2}}{\mathrm{N}_{\mathrm{A}}}$

Number of moles of third gas

$=\displaystyle \frac{\mathrm{n}_{3}}{\mathrm{N}_{\mathrm{A}}}$

If no loss of energy then

$\mathrm{P}_{1}\mathrm{V}_{1}+\mathrm{P}_{2}\mathrm{V}_{2}+\mathrm{P}_{3}\mathrm{V}_{3}=$

$PV$

$\displaystyle \frac{\mathrm{n}_{1}}{\mathrm{N}_{\mathrm{A}}}\mathrm{R}\mathrm{T}_{1}+\frac{\mathrm{n}_{2}}{\mathrm{N}_{\mathrm{A}}}\mathrm{R}\mathrm{T}_{2}+\frac{\mathrm{n}_{3}}{\mathrm{N}_{\mathrm{A}}}\mathrm{R}\mathrm{T}_{3}$

$\displaystyle =\frac{\mathrm{n}_{1}+\mathrm{n}_{2}+\mathrm{n}_{3}}{\mathrm{N}_{\mathrm{A}}}\mathrm{R}\mathrm{T}_{mix}$

$\implies \displaystyle \mathrm{T}_{mix}=\frac{\mathrm{n}_{1}\mathrm{T}_{1}+\mathrm{n}_{2}\mathrm{T}_{2}+\mathrm{n}_{3}\mathrm{T}_{3}}{\mathrm{n}_{1}+\mathrm{n}_{2}+\mathrm{n}_{3}}$ .

One

$kg$ of a diatomic gas is at a pressure of

$8\times 10^{4}\ \mathrm{N}/\mathrm{m}^{2}$ . The density of the gas is

$4$

$\mathrm{k}\mathrm{g}/\mathrm{m}^{3}$ . What is the energy of the gas due to its thermal motion ?

0%
$3\times 10^{4}\mathrm{J}$
0%
$5\times 10^{4}\mathrm{J}$
0%
$6\times 10^{4}\mathrm{J}$
0%
$7\times 10^{4}\mathrm{J}$

Match List I and List II

List-I List-II

a) P-V graph(T is constant) e) St. line cutting temp axis at - 273 $^{0}$

b) P-T graph(V is constant) f) Rectangular hyperbola

c) V-T graph(pressure P is constant) g) A st.line parallel to axis

d) PV- P garph h) St. line passing trhough origin (T is constant)

0%
a-g,b-e,c-h, d-f.
0%
a-h, b-f c-g, d-e.
0%
a-e, b-g,c-f, d-h.
0%
a-f, b-h, c-e,d-g.

Explanation

(a) P-V graph (T is constant):

$PV = nRT$
at constant temperature

$PV = constant$

$P \alpha \frac{1}{V}$
so it will be Rectangular hyperbola.

(b) P-T Graph (V is constant) :

$PV = nRT$
at constant volume

$P \alpha T$
pressure is proportional to T, so it will be straight line passing through origin.

c) V-T graph(P is constant) :

$PV = nRT$
at constant pressure

$V \alpha T$
volume is proportional to T(Kelvin),so it will be straight line passing through origin, but if temperature is taken in

$^{o}C$ , then it will cut temperature axis at

$-273 ^{o}C$

(d) PV - P Graph (T is constant) :

$PV = nRT$
at constant Temperature

$PV = constant$

$PV$ is constant for constant temperature,so it will be straight line parallel to P axis.

The amount of heat energy required to raise the temperature of

$1\ g$ of Helium at NTP, from

$T_1K$ to

$T_2K$ is -

0%
$\displaystyle \dfrac{3}{2}N_a k_B (T_2 - T_1)$
0%
$\displaystyle \dfrac{3}{4}N_a k_B (T_2 - T_1)$
0%
$\displaystyle \frac{3}{4}N_a k_B (\frac{T_2}{T_1})$
0%
$\displaystyle \frac{3}{8}N_a k_B (T_2 - T_1)$

One mole each of monatomic, diatomic and triatomic ideal gases (kept in three different containers) whose initial volume and pressure are same, each is compressed till their pressure becomes twice the initial pressure. Then which of the following statements is/are incorrect?

0%
If the compression is isothermal then their final volumes will be same
0%
If the compression is adiabatic, then final volume will be different
0%
If the compression is adiabatic, then triatomic gas will have maximum final volume
0%
If the compression is adiabatic, then monoatomic gas will have maximum final volume

Explanation

If the process is isothermal, then the volume has to change in order to compensate for the change in pressure.
If the compression is adiabatic, then the triatomic gas will have the maximum value because of the value of

$\gamma$
Since, the adiabatic gas law will be followed,

$P {V}^{\gamma}$ = C

$P_{1}V^{\gamma}_{1}=P_{2}V^{\gamma}_{2}\Rightarrow P_{0}V^{\gamma}_{0}=2P_{0}V^{\gamma}_{2}$

$V_{2}=\left ( \frac{1}{2} \right )^{\gamma}V_{0} \Rightarrow V_{2}=\dfrac{V_{0}}{2^{\gamma}}$
Monoatomic gas

$\gamma$ is min

$V_{2}$ is max.

Producers gas is a mixture of

0%
Carbon monoxide and nitrogen gas
0%
Carbon monoxide and hydrogen gas
0%
Carbon monoxide and water vapour
0%
Carbon monoxide and nitrous oxide

Atom of an element is electrically

0%
Positive
0%
Negative
0%
Neutral
0%
None of these

An ideal gas can be expanded from an initial state to a certain volume through two different processes (i)

$PV^2 = constant\:and\:(ii)P = KV^2$ where K is a positive constant. Based on the given situation, choose the correct statements

0%
Final temperature in (ii) will be greater than in (i)
0%
Final temperature in (ii) will be less than in (i)
0%
Total heat given to the gas in case (ii) is greater than in (i)
0%
Total heat given to the gas in case (ii) is less than in (i)

Explanation

Since, the question says that this is an ideal gas,
the molecules will always follow the ideal gas equation irrespective of the law that is used for the transformation.
Hence, we use P =

$\dfrac{nRT}{V}$
This gives us a relation between T and V that can show us the variation of T as the gas is expanded.
TV = C {for the first case}
T = C

${V}^{3}$ {for the second case}
Now, in both the cases,
V is increasing as expansion is occuring.
In first case T will decrease, whereas in the second case T will increase.

K.E. of molecular motion appears as:

0%
Pressure
0%
P.E.
0%
Temperature
0%
All of the above

Explanation

In the kinetic theory of gasses, increasing the temperature of a gas increases the average kinetic energy of the molecules, causing increased motion. This increased motion increases the outward pressure of the gas, an expected result from the ideal gas equation

$PV=NkT$

Hence, the kinetic energy of the molecular motion appears as temperature since it varies with change in temperature.

Hence, the correct option is

$C$

At change of state the kinetic energy of the molecules of a substances increases greatly.

0%
True
0%
False
0%
Either
0%
Neither

At pressure P and absolute temperature T a mass M of an ideal gas fills a closed container of volume V. An additional mass 2M of the same gas is added into the container and the volume is then reduced to

$\dfrac{v}{3}$ and the temperature to

$\dfrac{T}{3}.$ The pressure of the gas will now be :

0%
$\dfrac{P}{3}$
0%
$P$
0%
$3 P$
0%
$9 P$

Explanation

$M_{0}$ is molecular mass of the gas then for initial condition PV

$=\dfrac{M}{M_{0}}.RT ........(1)$
After 2M mass has been added

${P}'.\dfrac{V}{3}=\dfrac{3M}{M_{0}}.R.\frac{T}{3} .........(2)$
By dividing (2) and (1)

${P}'=3P$

A diatomic gas is filled inside a conducting cylinder. Now we push the piston slowly to make volume of gas half of initial. Pick correct statements

0%
Pressure of gas Increases because there is more average change in linear momentum of molecule in each collision
0%
Pressure force on side wall of container increased
0%
Pressure force on piston is increased
0%
More molecules collide with piston per unit time

A rigid container has a hole in its wall. When the container is evacuated, its weight is 100 gm. When someair is filled in it at 27C, its weight becomes 200 gm. Now the temperature of air inside is increased by

$\Delta$ T, the weight becomes 150 gm.

$\Delta$ T should be :

0%
$27^{\circ}$
0%
$\dfrac{27^{\circ}} {4}$ C
0%
$300^{\circ}$
0%
$327^{\circ}$

Explanation

Initially mass of air is 200 - 100 = 100gm. finally mass of air is 150 - 100 = 50 gm. As there is a hole in the wall, pressure inside the container will remain constant =

$P_{0}$

$PV = nRT \Rightarrow T \propto \dfrac{1} {n}$
as number of moles of gas is halved, the temperature should be doubled (in K)

$T_{i} = 300K$ So,

$T_{f} = 600 K \Rightarrow \Delta T = 300 K = 300^{\circ}C$

The relation between the internal energy

$U$ and adiabatic constant

$\gamma$ is

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$\displaystyle U=\frac{PV}{\gamma-1}$
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$\displaystyle U=\frac{PV^{\gamma}}{\gamma-1}$
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$\displaystyle U=\frac{PV}{\gamma}$
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$\displaystyle U=\frac{\gamma}{PV}$

Explanation

$PV =nRT$

$U =( N_0n)f(\dfrac{1}{2}kT)$ .....(1) where

f is the no of degrees of freedom,

$N_0$ is Avogadro's number

n is the no of moles

k is the Boltzman's constant

T is the absolute temp

f is related to

$\gamma$ by

$\gamma = \dfrac{2+f}{f}$ where

$\gamma$ is the adiabatic exponent.

$\therefore \gamma -1 = \dfrac{2+f}{f} -1=\dfrac{2}{f}$

$\Rightarrow f =\dfrac{2}{\gamma -1}$ .....(2)

Substituting f from(2) in (1)

$U =( N_0n)f(\dfrac{1}{2}kT)= ( N_0n)\dfrac{2}{\gamma -1}(\dfrac{1}{2}kT)=( N_0n)\dfrac{2}{\gamma -1}(\dfrac{1}{2}\dfrac{R}{N_0}T)=\dfrac{nRT}{\gamma -1}=\dfrac{PV}{\gamma -1}$

Energy of all molecules of a monatomic gas having a volume V and pressure P is

$\frac{3}{2} PV$ . The total translational kinetic energy of all molecules of a diatomic gas at the same volume and pressure is

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$\dfrac{1}{2} PV$
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$\dfrac{3}{2} PV$
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$\dfrac{5}{2} PV$
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$3 PV$

Explanation

The no of translational degrees of freedom for mono atomic or diatomic gas is same and is 3.
Hence, translational KE will be same at same temp for both gases.
Mono atmic gas has only translational degree of freedom.
Hence, translational kinetic energy for diatomic gas is

$\dfrac{3}{2}PV$

A gas mixture consists of

$2\:mol$ of oxygen and

$4\:mol$ of argon at temperature

$T$ . Neglecting all vibrational modes, the total internal energy of the system is

0%
$4RT$
0%
$15RT$
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$9RT$
0%
$11RT$

Explanation

$O_2$ molecule has 3 translational and 2 rotational degrees i.e total of 5 degrees of freedom.
Since Argon is a noble gas, it consists only of atoms and the energy is purely translational kinetic energy.
Total internal energy:

$U = 2 \times \dfrac{5}{2}RT + 4 \times \dfrac{3}{2}RT = 11RT$

A cylinder of capacity

$20\:L$ is filled with

$H_2$ gas. The total average kinetic energy of translatory motion of its molecules is

$1.5\times 10^5\:J$ . The pressure of hydrogen in the cylinder is

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$2\times 10^6\:N/m^2$
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$3\times 10^6\:N/m^2$
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$4\times 10^6\:N/m^2$
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$5\times 10^6\:N/m^2$

Explanation

The degree of freedom of a diatomic gas (

${H}_{2}$ ) is =5

out of which 3 are related to translatory motion and 2 are rotatory.

according to Maxwell's Law of Equipartition Energy total energy is equally shared among all degree of freedom.

if total energy is

${E}_{t}$ , then translatory motion energy

$E = \dfrac{3}{5}{E}_{t}$

$\dfrac{3}{5}{E}_{t} = 1.5 \times {10}^{5}$

${E}_{t} = 2.5 \times {10}^{5} J$

total energy of diatomic gas

${E}_{t} = \dfrac{5}{2}PV$

$P = \dfrac{2{E}_{t}}{5V}$

$V = 20 \times {10}^{-3}$ because

$1 {m}^{3} = 1000 L$

$P = \dfrac{2 \times 2.5 \times {10}^{5}}{5 \times 20 \times {10}^{-3}}$

solve for

$P$

$P = 5 \times {10}^{6} N/{m}^{2}$

answer is D.

According to Kinetic theory of gases, molecules are:

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perfectly inelastic particles in random motion
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perfectly elastic particles in random motion
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perfectly inelastic particles at rest
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perfectly elastic particles at rest

Explanation

Assumptions of Kinetic Theory of Gases

All gases are made up of molecules which are constantly and persistently moving in random directions.
The separation between the molecules is much greater than the size of molecules.
When a gas sample is kept in a container, the molecules of the sample do not exert any force on the walls of the container during the collision.
The time interval of collision between two molecules, and between a molecule and the wall is considered to be very small.
All the collisions between molecules and even between molecules and wall are considered to be elastic.
All the molecules in a certain gas sample obey Newton’s laws of motion.
If a gas sample is left for a sufficient time, it eventually comes to a steady state. The density of molecules and the distribution of molecules are independent of position, distance and time.

Hence, the correct option is

$(B)$

A box contains

$N$ molecules of a perfect gas at temperature

$T_1$ and pressure

$P_1$ . The number of molecules in the box is doubled keeping the total kinetic energy of the gas same as before. If the new pressure is

$P_2$ and temperature is

$T_2$ , then

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$P_2=P_1$ , $T_2=T_1$
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$P_2=P_1$ , $\displaystyle T_2=\frac{T_1}{2}$
0%
$P_2=2P_1$ , $T_2=T_1$
0%
$P_2=2P_1$ , $\displaystyle T_2=\frac{T_1}{2}$

Explanation

$P_1V=NkT_1$ ....(1)

$P_2V=2NkT_2$ .....(2)
Dividing (2) by (1)

$\dfrac{P_2}{P_1}= \dfrac{2T_2}{T_1}$ ....(3)
Since KE remains same,

$N \times \dfrac{3}{2}kT_1 = 2N \times \dfrac{3}{2}kT_2$

$\Rightarrow T_2 = \frac{T_1}{2}$ ....(4)
Substituting (4) in (3)

$\frac{P_2}{P_1}=1$

$\Rightarrow P_2= P_1$

Gaseous hydrogen contained initially under standard conditions in a sealed vessel of volume

$V= 5L$ was cooled by

$\Delta T = 55 K$ . The internal energy of the gas will change by

0%
$-255 J$
0%
$200 J$
0%
$250 J$
0%
$-200 J$

Explanation

Volume of vessel

$V = 5L = 0.005$

$m^3$

Standard conditions :

$P = 1$ atm

$= 1.01325 \times 10^5$ Pa

$T = 273.15 K$

Using Ideal gas equation :

$PV = nRT$

$\therefore$

$(1.01325 \times 10^5)\times 0.005 = nR \times 273.15$

$\implies nR = 1.8547$

Degree of freedom for diatomic gas

$f = 5$

Change in internal energy

$\Delta U = \dfrac{f}{2}nR\Delta T = \dfrac{5}{2} \times 1.8547 \times (-55) = -255$

$J$

Two containers of equal volume contain the same gas at pressures

$P_1$ and

$P_2$ and absolute temperatures

$T_1$ and

$T_2$ , respectively. On joining the vessels, the gas reaches a common pressure

$P$ and common temperature

$T$ . The ratio

$\displaystyle\frac{P}{T}$ is equal to

0%
$\displaystyle\frac{P_1}{T_1}+\frac{P_2}{T_2}$
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$\displaystyle\frac{P_1T_1+P_2T_2}{{(T_1+T_2)}^2}$
0%
$\displaystyle\frac{P_1T_2+P_2T_1}{{(T_1+T_2)}^2}$
0%
$\displaystyle\frac{P_1}{2T_1}+\frac{P_2}{2T_2}$

Explanation

$P_1V=n_1RT_1$

$P_2V=n_2RT_2$

where V is the volume of each vessel.

When the vessels are joined,

$P(2V)=(n_1 + n_2)RT$

$\therefore \dfrac{P}{T}= \dfrac{1}{2}\dfrac{(n_1 + n_2)R}{V}=\dfrac{1}{2}(\dfrac{P_1}{T_1} + \dfrac{P_2}{T_2})=\dfrac{P_1}{2T_1} + \dfrac{P_2}{2T_2}$

The ratio of kinetic energy to potential energy for solids is

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$\displaystyle \frac{Ek}{U}<1$
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$\displaystyle \frac{Ek}{U}>1$
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$Ek=U$
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$Ek>U$

The correct curve for a stable diatomic molecule is

A vessel of volume 4 litres contains a mixture of 8 g of O

$_2$ , 14 g of N

$_2$ and 22g of CO

$_2$ at 27

$^o$ C. The pressure exerted by the mixture is

0%
$5 \times 10^6 N/m^2$
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$6 \times 10^3 N/m^2$
0%
10 atmosphere
0%
$7.79 \times 10^5 N/m^2$

Explanation

Moles of

$O_2=\dfrac{8}{32}=\dfrac{1}{4}$

Moles of

$N_2=\dfrac{14}{28}=\dfrac{1}{2}$

Moles of

$CO_2=\dfrac{22}{44}=\dfrac{1}{2}$

Thus total moles in the mixture=

$\dfrac{1}{4}+\dfrac{1}{2}+\dfrac{1}{2}=\dfrac{5}{4}$

Using Ideal Gas Equation,

$PV=nRT$

$\implies P=\dfrac{nRT}{V}=\dfrac{\dfrac{5}{4}\times 8.3\times 300}{4\times 10^{-3}}N/m^2\approx 7.79\times 10^{5}N/m^2$

A vessel contains

$1$ mole of O

$_2$ (molar mass

$32 gm$ ) at a temperature

$T$ . The pressure is

$P$ . An identical vessel containing

$1$ mole of He (molar mass

$4 gm$ ) at a temperature

$2T$ has a pressure :

0%
P
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$\displaystyle \frac{P}{8}$
0%
2P
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8P

Explanation

Using

$PV=nRT$

$\therefore P_{He} = 2 P$ as temperature of He is doubled.

CBSE Questions for Class 11 Engineering Physics Kinetic Theory Quiz 5 - MCQExams.com

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Practice Class 11 Engineering Physics Quiz Questions and Answers

CBSE Questions for Class 11 Engineering Physics Kinetic Theory Quiz 5 - MCQExams.com

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Practice Class 11 Engineering Physics Quiz Questions and Answers

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