MCQ Questions for Class 11 Engineering Physics Kinetic Theory Quiz 17

An ideal gas is at a temperature

$T$ having molecules each of mass

$\mathrm { m } .$ If

$\mathrm { k }$ is the Boltzmann's constant and

$2 \mathrm { kT } / \mathrm { m } = 1.40 \times 10 ^ { 5 } \mathrm { m } ^ { 2 } / \mathrm { s } ^ { 2 } .$ Find the percentage of the fraction of molecules whose speed lie in the range

$324\mathrm { m } / \mathrm { s }$ to

$326\mathrm { m } / \mathrm { s } .$

0%
$0.52 \%$
0%
$0.43 \%$
0%
$0.21 \%$
0%
$0.14 \%$

The rate of a gaseous relation is given by the expressions K[A] [B].If the volume of reaction suddenly reduced to one fourth of the initial volume,the reaction rate relative to the original cases.

0%
$\dfrac { 1 }{ 6 }$
0%
$\dfrac { 1 }{ 8 }$
0%
$8$
0%
$16$

A classical model of a diatomic molecule is a springy dumbbell, as shown, where the dumbbell is free to rotate about axes perpendicular to the spring. At high temperature when vibrational degree of freedom is excited, the specific heat per mole at constant volume, is.

0%
$\dfrac32R$
0%
$\dfrac52R$
0%
$\dfrac72R$
0%
$\dfrac75R$

Kinetic energy of a gas depends upon it's:

0%
Equivalent mass
0%
atomic mass
0%
Molecular mass
0%
None of these

The molar heat capacity of water at constant pressure is 75 JK-1 mol -1. When 1.0 kJ of heat is supplied to 100 g of water which is free to expand to increases in temperature of water is

0%
6.6 K
0%
1.2 K
0%
2.4 K
0%
4.8 k

A diatomic ideal gas is heated at constant volume unitl its pressure is doubled- the molar heat capacity for the whole process is

0%
$\frac {23R}{5}$
0%
$\frac {19R}{5}$
0%
$\frac {13R}{5}$
0%
$\frac {19R}{6}$

An ideas monotonic gas is taken through cyclic process ABCA as given in figure, what will be the ratio of heat absorbed and released by gas during process?

0%
$\frac {12}{11}$
0%
$\frac {11}{10}$
0%
$\frac {10}{9}$
0%
$\frac {9}{8}$

Critical temperature of a gas obeying van der waals' equation is :

0%
$\dfrac{8a}{27Rb}$
0%
$\dfrac{a}{27b^2r}$
0%
$3b$
0%
$\dfrac{1}{27Rb}$

A mixture of ideal gases 7 kg of nitrogen and 11 Kg of

$CO_2$ then (Take

$\gamma$ for nitrogen and

$CO_2$ as 1.4 and 1.3 respectively)

0%
Equivalent molecular weight of the mixture is 36.
0%
Equivalent molecular weight of the mixture is 18.
0%
$\gamma$ for the mixture is 5/2
0%
$\gamma$ for the mixture is 47/35

One mole of an ideal monoatomic gas (initial temperature

${ T }_{ 0 }$ is made to go through the cycle abca shown in the fig. if U denotes the internal energy then choose the incorrect option

0%
${ U }_{ c }-{ U }_{ 0 }={ 15.5RT }_{ 0 }$
0%
${ U }_{ b }-{ U }_{ a }={ 6.5RT }_{ 0 }$
0%
${ U }_{ c }>{ U }_{ b }={ U }_{ a }$
0%
${ U }_{ c }-{ U }_{ b }=6{ RT }_{ 0 }$

$70$ calories of heat is required to raise the temperature of

$2$ moles of an ideal gas at constant pressure from

$30^0 C$ to

$35^0 C$ ( R = 2 cal/mol-^0C ). The amount of heat required to raise the temperature of the same gas through the same range at constant volume is

0%
$30 Calorie$
0%
$50 Calorie$
0%
$70 Calorie$
0%
$90 Calorie$

The speed of sound in oxygen

$(O_2)$ at a certain temperature is

$460 ms^{-1}$ .The speed of sound in helium (He) at the same temperature will be (Assume both gasses to be ideal)

0%
$460 \sqrt { \frac { 200 }{ 21 } } ms^{-1}$
0%
$500 \sqrt { \frac { 200 }{ 21 } } ms^{-1}$
0%
$650 \sqrt {2} ms^{-1}$
0%
$330 \sqrt {2} ms^{-1}$

The molar heat capacity of water at constant pressure

$P$ is

$75J{K}^{-1}$

${mol}^{-1}$ . When

$1.0kJ$ of heat is supplied to

$100g$ of water which is free to expand, the increase in temperature of water is:

0%
$1.2K$
0%
$2.4K$
0%
$4.8K$
0%
$6.6K$

A filled gas characterized by parameter P,V and T and another jar B filled with a gas with parameter 2P, V/4 and 2T. The ration of the number of molecules in jar A to those in jar B in-

0%
$1:1$
0%
$1:2$
0%
$2:1$
0%
$4:1$

An ideal monoatomic gas undergoes a process in which its internal energy

$U$ and density

$p$ vary as

$U/p=constant$ . The ratio of change in internal energy and the work done by the gas is

0%
$\frac { 3 }{ 2 }$
0%
$\frac { -2 }{ 3 }$
0%
$\frac { 1 }{ 3 }$
0%
$\frac { 3 }{ 5 }$

One mole of

$O_2$ gas is contained in a box of volume

$V = 2m^3$ at pressure

$P_0$ and temperature

$300 \,K$ . The gas is now heated to

$600 \,K$ and the molecules now get dissociated into oxygen atoms. The new pressure of the gas is

0%
$P_0$
0%
$2P_0$
0%
$4P_0$
0%
$8P_0$

The average kinetic energy per molecule of an ideal nonliner polyatomic gas at the room temperature

$T_0$ will be (k is boltzman constant, neglect vibrational modes)

0%
$\frac{3}{2} kT_0$
0%
$\frac{5}{2} kT_0$
0%
$3 kT_0$
0%
$\frac{1}{2} kT_0$

A diatomic gas of molecules weight

$40gm/mole$ is filled in a container at

$300K$ . It is moving at a velocity

$200m/s$ . If it is suddenly stopped, the rise in temperature of gas is:

0%
$\frac{60}{R}K$
0%
$\frac{100}{R}K$
0%
$\frac{320}{R}K$
0%
$\frac{90}{R}K$

An ideal gas of one mole is kept in a rigid container of negligible heat capacity. If

$25J$ of heat is supplied the gas temperature raises by

${2}^{o}C$ . Then the gas may be

0%
helium
0%
argon
0%
oxygen
0%
carbon dioxide

Three vessels of equal capacity have gases at the same pressure and temperature. The first vessel contains monoatomic gas, the second contains diatomic gas and third contains polyatomic gas. Which vessel contains more number of molecules?

0%
First
0%
Second
0%
Third
0%
All vessels contain same number of molecules

An ideal monoatomic gas of given mass is heated at constant pressure. In this process, the fraction of supplied heat energy used for the increase of the internal energy of the gas is

0%
$\dfrac { 3 }{ 8 }$
0%
$\dfrac { 3 }{ 5 }$
0%
$\dfrac { 3 }{ 4 }$
0%
$\dfrac {5 }{ 3 }$

$3$ mole of

$Ag$ is heated from

$300K$ to

$1000K$ . Calculate

$\Delta H$ when

$P = 1atm$ and

$C_p = 23 + 0.01T$ .

0%
$62 \ kJ/mol$
0%
$45 \ kJ/mol$
0%
$38 \ kJ/mol$
0%
$54 \ kJ/mol$

Explanation

$\displaystyle \Delta H = \int_{T_1}^{T_2}nCpdT = n\int^{1000}_{300}(23+0.01T)dT$

$= 3\left[23T + \dfrac{0.01T^2}{2}\right]^{1000}_{300}$

$=3\left[\dfrac{2\times 23\times 700 + 9100}{2}\right]$

$61950\ J/mol \simeq 62 \ kJ/mol$

$5$ moles of an ideal gas at

$100 \,K$ are allowed to undergo reversible compression till its temperature becomes

$200 \,K$ .
If

$C_v = 28 \,JK^{-1} mol^{-1}$ , calculate

$\Delta U$ and

$\Delta PV$ for this process.

$(R = 8.0 \,JK^{-1} mol^{-1})$

0%
$\Delta U = 14 \,kJ; \Delta (pV) = 4 \,kJ$
0%
$\Delta U = 14 \,kJ; \Delta (pV) = 18 \,kJ$
0%
$\Delta U = 2.8 \,kJ; \Delta (pV) = 0.8 \,kJ$
0%
$\Delta U = 14 \,kJ; \Delta (pV) = 0.8 \,kJ$

Explanation

Solution:- (A)

$\Delta U = 14 \,kJ; \Delta (pV) = 4 \,kJ$

$n = 5; T_i = 100 \,K; T_f = 200 \,K;$

$C_v = 28 \,J/mol \,K;$ Ideal gas

$\Delta U = nC_v \Delta T$

$= 5 \,mol \times 28 \,J/mol \,K \times (200 - 100)K$

$= 14,000 \,J = 14 \,kJ$

$\Rightarrow C_p = C_v + R = (28 + 8) J/mol \,K$

$= 36 \,J/mol \,K$

$\Rightarrow \Delta H = nC_p \Delta T = 5 \,mol \times 36 \,J/mol \,K \times 100 \,K$

$= 18000 \,J = 18 \,kJ$

$\Delta H = \Delta U + \Delta (PV)$

$\Rightarrow \Delta(PV) = \Delta H - \Delta U = (18 - 14) kJ = 4 \,kJ$

If the diatomic gas molecule is not rigid but has in addition a vibrational mode, then for this gas

0%
$C_v=\dfrac{7R}{2}$
0%
$C_p=\dfrac{9R}{2}$
0%
$\gamma =\dfrac{9}{7}$
0%
All of these

Consider a collision between an argon molecule and a nitrogen molecule in mixture of argon and nitrogen kept at room temperature. Which of the following are possible ?

0%
The kinetic energies of both the molecules decrese
0%
The kinetic energies of both the molecules increase
0%
The kinetic energy of the both the molecules increase
0%
The kinetic energy of the nitrogen molecules increases and that of the argon molecle decrease.

$\alpha$ moles of a monoatomic gas are mixed with

$\beta$ moles of a polyatomic gas and mixture behaves like diatomic gas, then [ neglect the vibrational mode of freedom )

0%
$2\alpha =\beta$
0%
$\alpha =2\beta$
0%
$\alpha =-3\beta$
0%
$3\alpha =-\beta$

An average human produces about

$10MJ$ of heat each day through metabolic activity. If a human body were an isolated system of mass

$80kg$ with the heat capacity of water, what temperature rise would be body experience? Heat capacity of water

$=4.2J/K-g$

0%
${ 29.76 } K$
0%
$2.976K$
0%
${ 2.976\times {{ 10 }^{ 4 } }}^{ o }C$
0%
${ 0.029 }^{ o }C\quad$

Explanation

heat produced

$=10MJ=10\times {10}^{6}J$

mass

$=80kg$

$C=4.2J/K-g$

we have relation

$Q=mc\Delta T$

$10\times {10}^{6}=80\times {10}^{3}\times 4.2\times \Delta T$

$\Delta T=\cfrac{10\times {10}^{6}}{80\times {10}^{3}\times 4.2}$

$\Delta T=0.02976\times {10}^{3}$

$\Delta T={29.76} K$

Option A is correct.

The total kinetic energy of all the molecules of helium having a volume V exerting a pressure P is 1500 J. The total kinetic energy in joules of all the molecules of

$N_{2}$ having the same volume V exerting a pressure 2P is-

0%
3000
0%
4000
0%
5000
0%
6000

A container has N molecules at absolute temperature T If the number of molecules is doubled but kinetic energy in box remains the same as before, the absolute temperature of the gas is

0%
T
0%
T/2
0%
2T
0%
zero

The molar heat capacity in a process of a diatomic gas if does a work of

$\frac{Q}{4}$ when a heat of Q is supplied to is

0%
$\frac{2}{5}R$
0%
$\frac{5}{2}R$
0%
$\frac{10}{3}R$
0%
$\frac{6}{7}R$

One mole of an ideal gas passes through a process where pressure and volume obey the relation

$P = P_0 \left[1 - \dfrac{1}{2} \left(\dfrac{V_0}{V} \right)^2 \right]$ . Here

$P_0$ and

$V_0$ are constants . Calculate the change in the temperature of the gas if its volume change from

$V_o$ to

$2 V_o$ .

0%
$\dfrac{1}{2} \dfrac{P_o V_o}{R}$
0%
$\dfrac{3}{4} \dfrac{P_o V_o}{R}$
0%
$\dfrac{5}{4} \dfrac{P_o V_o}{R}$
0%
$\dfrac{1}{4} \dfrac{P_oV_o}{R}$

An ideal gas is taken reversibly from state

$A(P,V)$ to the state

$B(0.5P,2V)$ along a straight line in PV diagram. Which of the following statement(s) is/are correct regarding the process?

0%
The work done by the gas in the process A to B exceeds the work that would be done by it if the same change in the state were performed isothermally
0%
In the T-V diagram, the path AB becomes a part of parabola
0%
In the P-T diagram, the path AB becomes a part of hyperbola
0%
Ongoing from A to B, the temperature of the gas first increases to the maximum value and the decreases

Three lawn chairs, one made up of aluminium (heat capacity

$=0.90J/K-g$ ), one of iron (heat capacity

$=0.45J/K-g$ ) and one of tin (heat capacity

$=0.60J/K-g$ ) are painted of the same colour. On a sunny day which chair will be hotter to sie?

0%
iron chair
0%
tin chair
0%
aluminium chair
0%
all, same

The values of

$C_{P, m}$ of

$N_{2}(g)$ and

$H_{2}O (g)$ (in

$cal/K-mol$ ) should be

0%
$8.3, 8.3$
0%
$8.3, 11.3$
0%
$11.3, 11.3$
0%
$11.3, 8.3$

If internal energy U of an ideal gas depends on pressure p and volume
V of the gas according to equation U = 3p V, which of the following
conclusions can you make regarding the gas?

0%
The gas is not mono-atomic gas
0%
The gas can be a di-atomic gas
0%
the gas can be tri-atomic gas
0%
Molar specific heat of the gas in isobaric process is 4 R

One gram-mole of an ideal gas

$A$ with the ratio of constant pressure and constant volume specific heats,

$\gamma _A = \dfrac{5}{3}$ is mixed with

$n$ gram-moles of another ideal gas

$B$ with

$\gamma_B = \dfrac{7}{5}$ . If the

$\gamma$ for the mixture is

$\dfrac{19}{13}$ what will be the value of

$n$ ?

0%
$0.75$
0%
$2$
0%
$1$
0%
$3$

The heat capacity of liquid water is

$75.6J/K-mol$ , while the enthalpy of fusion of ice is

$6.0kJ/mol$ . What is the smallest number of ice cubes at

${0}^{o}C$ , each containing

$9.0g$ of water, needed to cool

$500g$ of liquid water from

${20}^{o}C$ to

${0}^{o}C$ ?

0%
$1$
0%
$7$
0%
$14$
0%
$21$

Explanation

$500g$ of

${H}_{2}O$

moles of

${H}_{2}O=\cfrac{500g}{18g/mol}=27.78moles$

$Q$ heat generated

$=mc\Delta T$

$=27.78 mol \times 75.6 J/K- mol\times (293-273)K$

$=42,000J$

$Q=42,000\times \dfrac{1kJ}{1000J}=42kJ/mol$

Heat of fusion

$=6kJ/mol$

But we have

$9g$

${H}_{2}O$

moles

$=\cfrac{9g}{18g/mol}=0.5mol$

So,

$0.5$ mole of ice (1 ice cube) will have

$3\ kJ$ of energy.

500g of liquid water contains 42kJ of energy.

9g of water at

$0^oC$ = 1 ice cube contains 3kJ energy.

So, the number of ice cubes required

$=\dfrac{Total\ energy\ of\ liquid\ water}{energy\ per\ ice\ cube}=\cfrac{42}{3}=14$

Hence, the correct option is

$C$

In a model of chlorine

$(Cl_2)$ , two

$Cl$ atoms are rotated about their centre of mass as shown. Here the two

$'CI'$ atoms are

$2\times 10^{-10}m$ apart and angular speed

$\omega =2\times 10^{12}\ rad/s$ . If the molar mass of chlorine is

$70\ g/mol$ , then what is the rotational kinetic energy of one

$Cl_2$ molecule?

0%
$2.32\times 10^{-20}\ J$
0%
$2.32\times 10^{-21}\ J$
0%
$2.32\times 10^{-19}\ J$
0%
$2.32\times 10^{-22}\ J$

If at NTP, velocity of sound in a gas is

$1150\ m/s$ , then the

$rms$ velocity of gas molecules at

$NTP$ is :( Given :

$R=8.3\ joule/mol/K, C_p=4.8\ cal/mol/K$ )

0%
$1600\ m/s$
0%
$1532.19\ m/s$
0%
$160\ m/s$
0%
$16\ m/s$

We would like to increase the length of a

$15\ cm$ long copper rod of cross-section

$4\ mm^2$ by

$1\ mm$ . The energy absorbed by the rod if it is heated is

$E_1$ . The energy absorbed by the rod if it is stretched slowly is

$E_2$ . Then

$E_1 /E_2$ is:

0%
$300$
0%
$500$
0%
$480$
0%
$580$

Certain amount of an ideal gas is contained in a closed vessel. The vessel is moving with a constant velocity

$v$ The molecular mass of gas is

$M .$ The rise in temperature of the gas when the vessel is suddenly stopped is

$\left(\gamma=C_{P} / C_{V}\right)$

0%
$\dfrac{M v^{2}(\gamma-1)}{2 R(\gamma+1)}$
0%
$\dfrac{M v^{2}(\gamma-1)}{2 R}$
0%
$\dfrac{M v^{2}}{2 R(\gamma+1)}$
0%
$\dfrac{M v^{2}}{2 R(\gamma-1)}$

Explanation

$m$ is the total mass of the gas, then its kinetic energy

$=\dfrac{1 } {2} m v^{2}$ When the vessel is suddenly stopped, total kinetic energy will increase the temperature of the gas (because the process will be adiabatic), i.e.

$\dfrac{1}{2} m v^{2} =\mu C_{v} \Delta T$

$=\dfrac{m}{M} C_{v} \Delta T$

$\Rightarrow \dfrac{m}{M} \dfrac{R}{\gamma-1} \Delta T =\dfrac{1}{2} m v^{2} \quad\quad\quad\left(\mathrm{As} C_{v}=\dfrac{R}{\gamma-1}\right)$

$\Rightarrow \Delta T =\dfrac{M v^{2}(\gamma-1)}{2 R}$

An iron rocket fragment initially at

$-100^o$ C enters the earth's atmosphere almost horizontally and quickly fuses completely in atmospheric fiction. Specific heat of iron is 0.11 kcal/kg

$^o$ C its melting point is

$1535^o$ C and the latent heat of fusion is 3 kcals/kg. The minimum velocity with which the fragment must have entered the atmosphere is

0%
$0.45 km/s$
0%
$1.32 km/s$
0%
$2.32 km/s$
0%
$zero$

Explanation

The entire kinetic energy of the fragment is changed to heat. Expressing mass

$m$ in kilograms everywhere, we have

$Kinetic \ energy=\dfrac{1}{2}mv^2$

$gain \ energy=ms\Delta T+mL$

$\dfrac{1}{2} mv^2 = [m(30) + m(0.11)(1535^o + 100^o)]4184$

$v^2 = (8368) [30 + 180] = 1.76 \times 10^{6}$
Hence

$v = (\sqrt {1.76}) \times 10^3 = 1.32 km/s$

In a diatomic molecule, the rotational energy at a given temperature

0%
Obeys Maxwell's distribution
0%
Have the same value for all molecules
0%
Equals the translational kinetic energy for each molecule
0%
Is $\cfrac{2}{3}$ rd the translational kinetic energy for each molecule

Explanation

Consider a diatomic molecule along z-axis so its rotational energy about z-axis is zero. So energy of diatomic molecule

$E=\cfrac{1}{2}m{v}_{x}^{2}+\cfrac{1}{2}m{v}_{y}^{2}+\cfrac { 1 }{ 2 } m{v}_{z}^{2}+\cfrac { 1 }{ 2 } {I}_{x}{w}_{x}^{2}+\cfrac { 1 }{ 2 } {I}_{y}{w}_{y}^{2}$ (as moment of inertia along z-axis is zero)
The independent terms in the above expression is

$5$ . As we can predict velocities of molecules by Maxwell's distribution. Hence, the above expression also obeys Maxwell's distribution. As 2 rotational and 3 translational energies are associated with each molecule. So the rotational energy at given temperature is 2/3 of its translational Kinetic energy of each molecule.

That gas cannot be liquified

0%
Which obeys Vander Waal's equation
0%
Which obeys gas equation at every temperature and pressure
0%
The molecules of which are having potential energy
0%
Which is a inert gas

The kinetic energy , due to translational motion , of most of the molecules of an ideal gas at absolute temperature

$T$ is

0%
$kT$
0%
$\dfrac{k}{T}$
0%
$\dfrac{T}{k}$
0%
$\dfrac{1}{kT}$

CBSE Questions for Class 11 Engineering Physics Kinetic Theory Quiz 17 - 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 17 - MCQExams.com

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Practice Class 11 Engineering Physics Quiz Questions and Answers

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