At constant temperature the $$KE$$ is less for heavier gas molecules

At constant temperature the $$KE$$ of all the gas molecules is the same

At constant temperature, averge $$KE$$ of gas molecules remain constant but $$KE$$ of individual molecule may differ

At constant temperature the $$KE$$ is less for lighter gas molecules

MCQ Questions for Class 11 Engineering Physics Kinetic Theory Quiz 15

The rms speed of an ideal diatomic gas at temperature T is v . when gas dissociates into atoms then its new rms speed becomes double The temperature at which the gas dissociated into atoms are:-

0%
$T$
0%
$\sqrt{2}T$
0%
$\dfrac{T}{2}$
0%
$2T$

a container is filled with 20 moles of an ideal diatomic gas at an absolute temperature t when heat is supplied to gas temperature remains constant but 8 moles dissociate into the atom. heat energy given to gas is?

0%
$4\ RT$
0%
$6\ RT$
0%
$3\ RT$
0%
$RT$

Root mean square speed of the molecules of ideal gas is

$v$ . If pressure is increased two times at constant temperature, the

$rms$ speed will become:

0%
$2v$
0%
$\dfrac{v}{2}$
0%
$4v$
0%
$v$

Explanation

$v_{rms}=\sqrt{\dfrac{3RT}{M}}$

As T is constant

Though pressure is increased

$v_{rms}$ remains unchanged

Two perfect gases having masses

$m_{1}$ and

$m_{2}$ at temperature

$T_{1}$ and

$T_{2}$ are mixed without any loses of internal kinetic energy of the molecules. The molecular weights of the gases are

$M_{1}$ and

$M_{2}$ . What is the final temperature of the mixture?

0%
$\dfrac {m_{1}T_{1} + m_{2}T_{2}}{m_{1} + m_{2}}$
0%
$\dfrac {M_{1}T_{1} + M_{2}T_{2}}{M_{1} + M_{2}}$
0%
$\dfrac {\dfrac {m_{1}}{M_{1}} T_{1} + \dfrac {m_{2}}{M_{2}}T_{2}}{\left (\dfrac {m_{1}}{M_{1}} + \dfrac {m_{2}}{M_{2}}\right )}$
0%
$\dfrac {\dfrac {M_{1}}{m_{1}} T_{1} + \dfrac {M_{2}}{m_{2}}T_{2}}{\dfrac {M_{1}}{m_{1}} + \dfrac {M_{2}}{m_{2}}}$

A particle of mass m is moving along the x-axis with speed v when it collides with a particle of mass

$2$ M initially at rest. After the collision, the first particle has come to rest and the second particle has split into two equal mass pieces that are shown in the figure. Which of the following statements correctly describes the speeds of the piece?

$(\theta > 0)$ .

0%
Each piece moves with speed v
0%
Each piece moves with speed $v/2$
0%
One of the pieces moves with speed $v/2$ , the other with speed greater than $v/2$
0%
Each piece moves with speed greater than $v/2$

Explanation

Let

${ V }_{ A }^{ / }$ and

${ V }_{ B }^{ / }$ be the velocity of A and B after collision.

From the conservation of momentum,

$\quad { M }_{ A }{ V }_{ A }+{ M }_{ B }{ V }_{ B }={ M }_{ A }{ V }_{ A }^{ / }+\quad { M }_{ B }^{ / }{ V }_{ B }^{ / }$ -------- (1)

But,

${ M }_{ B }=2m\quad$ and

${ M }_{ A }=m$

From (1), => mv+ 0= 0+ 2(

$\frac { 1 }{ 2 } \times 2m$ )v'Cos

$\theta$

=> v=2v'Cos

$\theta$

v'=

$\frac { v }{ 2Cos\theta } =\frac { 1 }{ Cos\theta } (\frac { v }{ 2 } )$

Hence each piece moves with speed greater than

$\dfrac{v}{2}$

The number of molecules in 1 cc of water is closed to

0%
6 X $10^{23}$
0%
22.4 X $10^{24}$
0%
$\frac{10^{23}}{3}$
0%
$10^{23}$

A closed cylindrical vessel contains

$N$ moles of an ideal diatomic gas at a temperature

$T$ . On supplying heat, the temperature remains same, but

$n$ moles get dissociated into atoms. The heat supplied is:

0%
$\dfrac{5}{2}(N - n)RT$
0%
$\dfrac{5}{2}n RT$
0%
$\dfrac{1}{2}n RT$
0%
$\dfrac{3}{2}n RT$

Explanation

Solution:- (C)

$\cfrac{1}{2} nRT$

Since the gas is enclosed in a vessel, during heating, volume of the gas remains constant.

$\therefore W = 0$

$\therefore q = \Delta{U} ..... \left( 1 \right)$

Now,

As we know that,

$\Delta{U} = n {C}_{v} \Delta{T}$

Initial internal energy of the gas is

${U}_{1} = N \left( \cfrac{5}{2} R \right) T = \cfrac{5}{2} NRT$

Since

$n$ moles dissociate into atoms, therefore after heating, the vessel contains

$\left( N - n \right)$ moles of diatomic gas and

$2n$ moles of mono atomic gas. Hence the internal energy of the gas after heating will be equal to

${U}_{2} = \left( N - n \right) \left( \cfrac{5}{2} R \right) T + 2n \left( \cfrac{3}{2} R \right) T$

$\Rightarrow { U}_{2} = \cfrac{1}{2} nRT + \cfrac{5}{2} NRT$

$\therefore \Delta{U} = {U}_{2} - {U}_{1}$

$\Rightarrow \Delta{U} = \left( \cfrac{1}{2} nRT + \cfrac{5}{2} NRT \right) - \cfrac{5}{2} NRT$

$\Rightarrow \Delta{U} = \cfrac{1}{2} nRT$

Now from

${eq}^{n} \left( 1 \right)$ , we get

$q = \cfrac{1}{2} nRT$

The pressure of an ideal gas veries according to the law

$P=P_{0}-AV^{2}$ where

$P_{0}$ and A are positive constants. What is the highest temperature that can be attained by the gas?

0%
$\dfrac{P_{0}}{nR}\sqrt{\dfrac{P_{0}}{A}}$
0%
$\dfrac{P_{0}}{nR}\sqrt{\dfrac{P_{0}}{2A}}$
0%
$\dfrac{2P_{0}}{nR}\sqrt{\dfrac{P_{0}}{2A}}$
0%
$\dfrac{2P_{0}}{3nR}\sqrt{\dfrac{P_{0}}{3A}}$

The heat capicity of liquid water at constant pressure,

$C_p$ is

$18$ cals

$deg^{-1} mol^{-1}$ . The value of heat capacity of water at constant volume, Cv is approximately:

0%
$18 cal deg^{-1} mol^{-1}$
0%
$16 cal deg^{-1} mol^{-1}$
0%
$10.8 cal deg^{-1} mol^{-1}$
0%
Cannot be predicted

$C_{\upsilon }$ values for monoatomic and diatomic gases respectively are:

0%
$\cfrac{R}{2},\cfrac{3R}{2}$
0%
$\cfrac{3R}{2},\cfrac{5R}{2}$
0%
$\cfrac{5R}{2},\cfrac{7R}{2}$
0%
$\cfrac{3R}{2},\cfrac{3R}{2}$

A diatomic gas is undergoing a process for which P versus V relation is given as

${ PV }^{ -\frac { 5 }{ 3 } }$ = constant. The molar heat capacity of the gas for this process is:

0%
$\cfrac{5R}{3}$
0%
$\cfrac{23R}{8}$
0%
$\cfrac{17R}{4}$
0%
$\cfrac{13R}{5}$

Two closed containers of equal volume filled with air at pressure

$P_{0}$ and temperature

$T_{0}$ . Both are connected by narrow tube. If one of the container is maintained at temperature

$T_{0}$ and another at temperature T, then new pressure in the containers will be

0%
$\dfrac{2P_{0}T}{T+T_{0}}$
0%
$\dfrac{P_{0}T}{T+T_{0}}$
0%
$\dfrac{P_{0}T}{2(T+T_{0}})$
0%
$\dfrac{T+T_{0}}{P_{0}}$

The molar heat capacity of eater at constant pressure, C, is

$75JK^{-1}mol^{-1}$ . When 1.0KJ of heat is supplied to 100g of water which is free to expand, the increase in temperature of water is :

0%
0.24K
0%
2.4K
0%
1.3K
0%
0.13K

The heat capacity of liquid water is 75.6 J/mol k, while the enthaply of fusion of ice is 6.0 kJ/mol. What is the smallest number of ice cubes at

$0^{0}C$ , each containing 9.0 g of water neede to cool 500 g of liquid water from

$20^{0}C$ to

$0^{0}C$ ?

0%
1
0%
7
0%
14
0%
None of these

A gas has molar heat capacity

$C = 4.5\ R$ in the process

$PT = constant$ . Find the number of degrees of freedom (n) of molecules in the gas.

0%
$n = 7$
0%
$n = 3$
0%
$n = 5$
0%
$n = 2$

A particle moving along x-axis is acted upon by a single force

$F=F_0e^{-kx}$ , where

$F_0$ and K and constants. The particle is released from rest at x=It will attain a maximum KE of

0%
$\frac{F_0}{K}$
0%
$\frac{F_0}{e^K}$
0%
$KF_0$
0%
$Ke^KF_0$

Figure shows a hypothetical speed distribution for particles of a certain gas

$P ( v ) = C r ^ { 2 } \text { for } 0 < v \leq v _ { 0 } \text { and } P ( v ) = 0 \text { for } v > V _ { 0 }$

0%
$C = 3 / V ^ { 3 }$
0%
average speed = $0.750 \mathrm { V } _ { 0 }$
0%
rms speed= $0.775 \mathrm { V } _ { 0 }$
0%
$C = 3 / V ^ { 2 }$

A gas of molecule of mass M at the surface of the Earth has kinetic energy equivalent to

${0^0}C$ . If it were to go up straight without colliding with any other molecules, how high it would rise? Assuming that the height attained is much less than radius of the earth.

0%
$0$
0%
$\frac{{273{k_B}}}{{2{M_g}}}$
0%
$\frac{{546{k_B}}}{{3{M_g}}}$
0%
$\frac{{819{k_B}}}{{2{M_g}}}$

N moles of an ideal diatomic gas is contained in a cylinder at temperature

$T.$ On supplying some heat to cylinder,

$N/3$ moles of gas disassociated into atoms while temperature remains constant. Heat supplied to the gas is

0%
$\dfrac {NRT}{3}$
0%
$\dfrac {5NRT}{2}$
0%
$\dfrac {8NRT}{3}$
0%
$\dfrac {NRT}{6}$

The respective speeds of the molecules are

$1,2,3,4$ and

$5\,km, {s}^{-1}$ . The ratio of their r.m.s. velocity and the average velocity will be

0%
$\sqrt{11} : 3$
0%
$3 : \sqrt{11 }$
0%
$1 :2$
0%
$3 : 4$

A box contains N molecules of a perfact 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

$T_{2}$ , then

0%
$P_{2}=P_{1},T_{2}=T_{1}$
0%
$P_{2}=P_{1},T_{2}=\cfrac{T_{1}}{2}$
0%
$P_{2}=2P_{1},T_{2}=T_{1}$
0%
$P_{2}=2P_{1},T_{2}=\cfrac{T_{1}}{2}$

One mole of an ideal gas undergoes a cyclic change as shown in the figure. The process AB is isothermal. The pressure and volume at point A is $2.1 \times {10^5}$ and 22.4 liters, respectively. The temperature at C is

0%
${10^0}{\text{C}}$
0%
273 K
0%
0 K
0%
${27^0}{\text{C}}$

The pressure of the gas contained in a closed vessel is increased by

$0.4\%$ when heated by

$1^oC$ . The initial temperature of the gas must be

0%
$1250 K$
0%
$250 K$
0%
$2500 K$
0%
$25^oC$

Energy associated with each molecule per degree of freedom of a system at room temperature (

$27^{o}C$ ) will be (

$k$ is Boltzmann;s constant)

0%
$150\ k$
0%
$(27/2)\ k$
0%
$1/2\ R$
0%
$none\ of\ these$

For a particle moving along x-axis, acceleration is given as

$a=2v^2$ . If the speed of the particle is

$v_0$ at

$x=0$ , find speed as a function of

$x$ .

0%
$v_0e^{2x}$
0%
$v_0e^{-2x}$
0%
$v_0e^{-4x}$
0%
$v_0e^{4x}$

70 calorie of heat required to rise the temperature of 2 mole of an ideal gas at constant pressure from

${30^o}$ C to

${35^o}$ C. The degrees of freedom of the gas molecule are,,

0%
3
0%
5
0%
6
0%
7

If the total energy of a

$H_2$ gas molecule is x, the linear energy of the gas molecule is

0%
X
0%
$\dfrac{3x}{2}$
0%
$\dfrac{3x}{5}$
0%
$\dfrac{5x}{3}$

Three perfect gases at absolute temperatures

$T_1, T_2$ and

$T_3$ are mixed. The masses of their molecules are

$m_1, m_2$ and

$m_3$ and the number of molecules are

$n_1, n_2$ and

$n_3$ respectively. Assuming no loss of energy, the final tempreture of the mixture is

0%
$\dfrac{T_1 + T_2 + T_3}{3}$
0%
$\dfrac{n_1T_1 + n_2T_2 + T_3 T_3}{n_1 + n_2 + n_3}$
0%
$\dfrac{n_1T_1^2 + n_2T_2^2 + n_3 T_3^2}{n_1 T_1 + n_2 T_2 + n_3 T_3}$
0%
$\dfrac{n_1^2T_1^2 + n_2^2T_2^2 + n_3^2 T_3^2}{n_1 T_1 + n_2 T_2 + n_3 T_3}$

A horizontal insulated cylinder of volume

$V$ is divided into four identical compartments by stationary semi-permeable thin partitions as shown. The four compartments from left are initially filled with 28 g helium, 160 g oxygen, 28 g nitrogen and 20 g hydrogen respectively. The left partition lets through hydrogen, nitrogen and helium while the right partition lets through hydrogen only. The middle part lets through hydrogen and nitrogen both. The temperature

$T$ inside the entire cylinder is maintained constant. After the system is set in equilibrium

0%
pressure of helium $\frac { 14 R T } { V }$
0%
pressure of oxygen $\frac { 20 R T } { V }$
0%
pressure of nitrogen $\frac { 4 R T } { V }$
0%
pressure of hydrogen $\frac { 10 R T } { V }$

A mass of liquid with volume V, completely turns into a gas of volume

$V_2$ at a constant external pressure P and temperature T. The latent heat of vapourisation is L for the given mass. Then change in the internal energy of the system is

0%
$PV_2-PV_1$
0%
$L-PV_2+PV_1$
0%
Zero
0%
$L+PV_2-PV_1$

A jar contains a mixture of oxygen and helium gases in the ratio 64 :Ratio of their mean translational energies at same temperature is

0%
1 : 1
0%
1 : $\sqrt 2$
0%
2 : 1
0%
1 : 2

Correct statement

0%
At constant temperature the $KE$ of all the gas molecules is the same
0%
At constant temperature, averge $KE$ of gas molecules remain constant but $KE$ of individual molecule may differ
0%
At constant temperature the $KE$ is less for heavier gas molecules
0%
At constant temperature the $KE$ is less for lighter gas molecules

A vessel contains a non-linear triatomic gas. If

$50$ % of gas dissociate into individual atom, then find new value of degree of freedom by ignoring the vibrational mode and any further dissociation.

0%
2.15
0%
3.75
0%
5.25
0%
6.35

one mole of mono atomic ideal gas follows a process AB, as shown. The specific heat of the process is

$\dfrac{13R}{6}$ . Find the value of X on P-axis.

0%
$4P_0$
0%
$5P_0$
0%
$6P_0$
0%
$8P_0$

A particle located at

$x=0$ at time

$t=0$ , starts moving along the positive x-direction with a velocity

$v$ tha varies as

$v=\alpha \sqrt{x}$ , the displacement of the particle varies with time as

0%
${t}^{2}$
0%
$t$
0%
${t}^{1/2}$
0%
${t}^{3}$

An electric bulb of volume

$250\ cm^{3}$ has been sealed at a pressure of

$10^{-3} mm$ of mercury and temperature

$27^{\circ}C$ . Find the number of air molecules in the bulb. What is the average distance between the molecules?

0%
$9.02\times 10^{15}, 3.15\times 10^{-5} m$ .
0%
$8.02\times 10^{10}, 3.15\times 10^{-6} m$ .
0%
$4.02\times 10^{15}, 2.15\times 10^{-7} m$ .
0%
$8.02\times 10^{15}, 3.15\times 10^{-7} m$ .

$5\ mol.$ of oxygen is heated at constant volume from

$10^{\circ}C$ to

$20^{\circ}C$ . The change in the internal energy of the gas is

0%
$220\ calories$
0%
$240\ calories$
0%
$252\ calories$
0%
$300\ calories$

For one mole of monoatomic gas, work done at constant pressure is W. The heat supplied at constant volume for the same rise in temperature of the gas is

0%
W/2
0%
3W/2
0%
5W/2
0%
W

A diatomic gaas molecular weight 30 gm/ mole is filled in a container at 27

${ \circ }_{ C }$ . It is moving at a velocity 100 m/s . If it is suddenly stopped , the rise in temperature of gas is :

0%
60/R
0%
$\dfrac { 600 }{ R }$
0%
$\dfrac { 600\times { 10 }^{ 4 } }{ R }$
0%
$\dfrac { 6\times { 10 }^{ 5 } }{ R }$

A diatomic gas is heated at constant pressure. What fraction of the heat energy is used to increase the internal energy ?

0%
$\frac 3 5$
0%
$\frac 3 7$
0%
$\frac 5 7$
0%
$\frac 5 9$

$_{0}$ and E

$_{H}$ respectively represents the average kinetic energy of a molecule of oxygen and hydrogen .If the two gases are at the same temperature , which of the following statement are true?

0%
E $_{0}$ > E $_{H}$
0%
E $_{0}$ = E $_{H}$
0%
E $_{0}$ < E $_{H}$
0%
Data insufficient

Heat is supplied to a diatomic gas at constant pressure . The ratio of

$\triangle Q : \triangle U : \triangle W$ is

0%
5 : 3 : 2
0%
7 : 5 : 2
0%
2 : 3 : 5
0%
2 : 5 : 7

One mole of an mono atomic gas is taken round the cyclic process ABCA as shown in figure, then

0%
Work done on the gas is ${ P }_{ 0 }{ V }_{ 0 }$
0%
Heat absorbed in the process CA is $5/2{ P }_{ 0 }{ V }_{ 0 }$ and rejected in the process AB is ${ P }_{ 0 }{ V }_{ 0 }$
0%
Heat rejected in the process BC
0%
$\frac { { P }_{ 0 }{ V }_{ 0 } }{ R }$ is the maximum temperature attained by the gas during cycle

$4.0g$ of a gas occupies

$22.4 litres$ at

$NTP$ . The specific heat capacity of the gas at constant volume is

$5.0{ JK }^{ -1 }{mol }^{ -1 }$ . If the speed of sound in this gas at NTP is

$952ms^{-1}$ , then the heat capacity at constant pressure is(Take gas constant

$R= 8.3{ JK }^{ -1 }{mol }^{ -1 }$ )

0%
$8.5{ JK }^{ -1 }{mol }^{ -1 }$
0%
$8.0{ JK }^{ -1 }{mol }^{ -1 }$
0%
$7.5{ JK }^{ -1 }{mol }^{ -1 }$
0%
$7.0{ JK }^{ -1 }{mol }^{ -1 }$

The molecules of an ideal gas have

$6degrees$ of freedom. The temperature of the gas is

$T$ . The average translational kinetic energy of its molecules is:

0%
$\frac {3}{2}kT$
0%
$\frac {6}{2}kT$
0%
$kT$
0%
$\frac {1}{2}kT$

A perfect gas at

$27^0C$ is heated at constant pressure so as to triple its volume. The temperature of the gas will be

0%
$81^0C$
0%
$900^0C$
0%
$627^0C$
0%
$450^0C$

Two blocks of the same metal having the same mass and at temperature

$T_1$ and

$T_2$ , respectively. are brought in contact with each other and allowed to attain thermal equilibrium at constant pressure. The change in entropy,

$\Delta S$ , for this process is :

0%
$2 C_p ln \left(\dfrac{T_1 + T_2}{4T_1 T_2} \right)$
0%
$2 C_p ln \left[\dfrac{(T_1 + T_2)^{\dfrac{1}{2}}}{T_1 T_2} \right]$
0%
$C_p ln \left[\dfrac{(T_1 + T_2)^2}{4T_1 T_2} \right]$
0%
$2C_p ln \left[\dfrac{T_1 + T_2}{2T_1 T_2} \right]$

Explanation

When two blocks are kept in contact with each other, the final temperature will be given as:

$T_f=\dfrac{T_1+T_2}{2}$

Now, we have

$\Delta S_{sys}=\int \dfrac{dq_{rev}}{T}=nC_p\int \dfrac{dT}{T}$

For the first block of metal the entropy will be given as:

$\Delta S_1=nC_p \int ^{T_f}_{T_1} \dfrac{dT}{T}=nC_pln\dfrac{T_f}{T_1}$

Similarily,

$\Delta S_2=nC_p ln \dfrac{T_f}{T_2}$

Now, total change in entropy

$\Delta S_1+ \Delta S_2 =nC_p \dfrac{{T_f}^2}{T_1 T_2}$

But ,

$T_f=\dfrac{T_1+T_2}{2}$

$\therefore$ Final entropy will be:

$nC_p ln [\dfrac{(T_1+T_2)^2}{4T_1T_2}]$

An ideal gas in a container of volume

$500c.c.$ is at a pressure of

$2\times 10^5Nm^{-2}$ . The average kinetic energy of each molecule

$6\times 10^{-21}J$ . The number of gas molecules in a container is:

0%
$25\times 10^{25}J$
0%
$5\times 10^{23}J$
0%
$25\times 10^{23}J$
0%
$2.5\times 10^{22}J$

One mole of monoatomic gas is carried along process ABCDEA as shown in the diagram.Find the net work done by gas.

0%
$\frac {3}{2} J$
0%
1 J
0%
$\frac {1}{2} J$
0%
0 J

At 323 K, the vapour pressure in millimeters of mercury of a methanol-ethanol solution is represented by the equation

$p=120{ x }_{ A }+140$ ,where

${ x }_{ A }$ is the mole fraction of methanol. Then the value of

$\lim _{ { X }_{ A }\rightarrow 1 }{ \frac { { P }_{ A } }{ { X }_{ A } } }$

0%
250 mm
0%
140 mm
0%
260 mm
0%
20 mm

CBSE Questions for Class 11 Engineering Physics Kinetic Theory Quiz 15 - 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 15 - MCQExams.com

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Practice Class 11 Engineering Physics Quiz Questions and Answers

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