MCQ Questions for Class 11 Engineering Physics Kinetic Theory Quiz 10

Mean kinetic energy(or average energy) per gm molecule of a monoatomic gas is given by:

0%
3RT/2
0%
kT/2
0%
RT/2
0%
3kT/2

Explanation

One gram molecule is the no.of molecules present in one mole, i.e,

$6.022 \times 10^23$ molecules.

Therefore mean kinetic energy per gram molecule of a monoatomic gas:

$=\dfrac{3}{2} Nk_B T$

$=\dfrac{3}{2} RT$ since

$R=Nk_B$

The de-Broglie wavelength of a particle accelerated with

$150\ volt$ potential is

$10^{-10}\ m$ . If it accelerated by

$600\ volts$ p.d. its wavelength will be

0%
$0.25\ A^{o}$
0%
$0. 5\ A^{o}$
0%
$1.5\ A^{o}$
0%
$2\ A^{o}$

Explanation

Given,

$\lambda =\dfrac{hc}{eV}\ \ \ \ where,\ V=potential$

$\lambda \ \alpha \ \dfrac{1}{V}$

${{10}^{-10}}\ \alpha \ \dfrac{1}{150}\ ......\ (1)$

$\lambda \ \alpha \ \dfrac{1}{600}\ ......\ (2)$

Divide (2) by (1)

$\dfrac{\lambda }{{{10}^{-10}}}=\dfrac{150}{600}=\dfrac{1}{4}$

$\Rightarrow \lambda =0.25\times {{10}^{-10}}m\ =0.25\ {{A}^{o}}$

We have a jar A filled with gas characterized by parameters P, V and T and another jar B filed with gas with parameters 2P, Vl4 and 2T, where the symbols have their usual meanings. The ratio of the number of molecules of jar A to those of jar B is

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

$\triangle E$ is the heat of reaction for

${ C }_{ 2 }{ H }_{ 3 }{ OH }_{ \left( 1 \right) }+{ 3O }_{ 2\left( g \right) }\rightarrow { 2CO }_{ 2\left( s \right) }+3{ H }_{ 2 }{ 0 }_{ \left( 1 \right) }$ at constant volume, the

$\triangle H$ ( Heat of reaction at constant pressure) at constant temperature is:

0%
$\triangle H=\triangle E+2RT$
0%
$\triangle H=\triangle E-2RT$
0%
$\triangle H=\triangle E+RT$
0%
$\triangle H=\triangle E-RT$

Explanation

$DH=DE+Dn_{gas} RT$

$\Delta n_{brq}=3-1$ (as a question)

$\therefore \ DH=DE+2RT$

At what temperature does the average translational kinetic energy of a molecule in a gas become equal to kinetic energy of an electron accelerated from rest through a potential difference of

$1 \ volt$ ?

$(k=1.38\times10^{23} \ J/k)$

0%
$3770K$
0%
$7370K$
0%
$7730K$
0%
$7330K$

The total kinetic energy of

$1$ mole of

$N_2$ at

$27$ C will be approximately

0%
3739.662 J
0%
1500 calorie
0%
1500 kilo calorie
0%
1500 erg.

Explanation

The kinetic enrgy of one mole is given by:

KE=

$\dfrac{3}{2} K_BT$

The kinetic enrgy of 1 mole of

$N_2$ atoms is:

KE=

$\dfrac{3}{2}K_B T$ where

$N$ is Avogadro's number,

$K_B$ is Boltzmann's constant and

$T$ is temperature

KE=

$\dfrac{3}{2} \times (6.022 \times 10^{23})\times (1.38 \times 10^{-23}) \times 300$

$=3739.662 J$

Three perfect gasses at absolute temperatures

$T_1$ ,

$T_2$ and

$T_3$ are mixed. If number of molecules of the gasses are

$n_1$ ,

$n_2$ and

$n_3$ respectively then temperature of mixture will be (assume no loss of energy)

0%
$\dfrac{ T_{1}+ T_{2}+T_{3}}{3}$
0%
$\dfrac{ { n }_{ 1 }^{ 2 } T_{1}+ { n }_{ 2 }^{ 2 } T_{2}+ { n }_{ 3 }^{ 2 } T_{3}}{{ n }_{ 1 }+{ n }_{ 2 }+{ n }_{ 3 }}$
0%
$\dfrac{ { n }_{ 1 } T_{1}+ { n }_{ 2 }T_{2}+ { n }_{ 3 }T_{3}}{{ n }_{ 1 }+{ n }_{ 2 }+{ n }_{ 3 }}$
0%
$\dfrac{ T_{1}+ T_{2}+T_{3}}{{ n }_{ 1 }+{ n }_{ 2 }+{ n }_{ 3 }}$

Explanation

For perfect gas,

Kinetic Energy of n molecule, $K.E=n\left( \dfrac{1}{2}{{K}_{B}}T \right)$

Where, ${{K}_{B}}$ is Boltzmann constant

If there is no loss of energy.

Total kinetic energy of mixture is sum of each gas kinetic energy.

${{n}_{total}}K.{{E}_{total}}={{n}_{1}}K.{{E}_{1}}+{{n}_{2}}K.{{E}_{2}}+{{n}_{3}}K.{{E}_{3}}$

$\left( {{n}_{1}}+{{n}_{2}}+{{n}_{3}} \right)\left( \dfrac{1}{2}{{K}_{B}}T \right)={{n}_{1}}\left( \dfrac{1}{2}{{K}_{B}}{{T}_{1}} \right)+{{n}_{2}}\left( \dfrac{1}{2}{{K}_{B}}{{T}_{2}} \right)+{{n}_{3}}\left( \dfrac{1}{2}{{K}_{B}}{{T}_{3}} \right)$

$T=\dfrac{{{n}_{1}}{{T}_{1}}+{{n}_{2}}{{T}_{2}}+{{n}_{3}}{{T}_{3}}}{{{n}_{1}}+{{n}_{2}}+{{n}_{3}}}$

Calculate

$\gamma$ (ratio of

$C_p$ and

$C_v$ ) for triatomic linear gas at high temperature. Assume that the contribution of the vibrational degree of freedom is 75%?

0%
1.222
0%
1.121
0%
1.18
0%
1.33

Explanation

For Triatomic molecule

$N=3$

Degree of freedom

$=3N=9$

$\longrightarrow$ Translation degree of freedom

For linear, 2

$\longrightarrow$ Rotational degree of freedom

$9-(3+2)=4\longrightarrow$ Vibrational degree of freedom

$U=3\times \cfrac { 1 }{ 2 } kT+2\times \cfrac { 1 }{ 2 } kT+4\times \cfrac { 75 }{ 100 } kT$

as vibrational contributes only 75%

$U=\cfrac { 3 }{ 2 } kT+kT+3kT=\cfrac { 11 }{ 2 } kT$

${ C }_{ V }=\cfrac { dU }{ dT } =\cfrac { 11 }{ 2 } k$ or for 1 mole

$\cfrac{11}{2}R$

As for ideal gas,

${ C }_{ P }-{ C }_{ V }=R\\ { C }_{ P }={ C }_{ V }+R=\cfrac { 11 }{ 2 } R+R=\cfrac { 13 }{ 2 } R\\ g=\cfrac { { C }_{ P } }{ { C }_{ V } } =\cfrac { 13 }{ 2 } R\times \cfrac { 2 }{ 11R } =1.18$

Thus, the ratio of

$C_{p}$ and

$C_{v}$ is 1.18.

A vessel has

$6g$ of oxegen at pressure

$P$ and temperature

$400\ K$ . A small hole is made in it so that oxygen leaks out. How much oxygen leaks out if the final pressure is

$P/2$ and temperature is

$300\ K$

0%
$3g$
0%
$2g$
0%
$4g$
0%
$5g$

Explanation

From ideas gas equation

$PV=nRT$

$PV=\dfrac{m}{{{M}_{o}}}RT$

$m=\dfrac{PV{{M}_{o}}}{RT}\$

In first event

${{m}_{1}}=\dfrac{{{P}_{1}}V{{M}_{o}}}{R{{T}_{1}}}$

$6\ gram=\dfrac{PV{{M}_{o}}}{R\times 400}\$

$\dfrac{{{\operatorname{PVM}}_{o}}}{R}=400\times 6\ .\ .....\ (1)$

In the second event when hole is made:

${{m}_{2}}=\dfrac{{{P}_{2}}V{{M}_{o}}}{R{{T}_{2}}}$

${{m}_{2}}=\dfrac{\dfrac{P}{2}V{{M}_{o}}}{R\times 300}\ =\dfrac{1}{300\times 2}\dfrac{PV{{M}_{o}}}{R}$

${{m}_{2}}=\dfrac{1}{300\times 2}\times \left( 400\times 6 \right)=4\ gram$

Leak out oxygen is $\left( 6gram-4\,gram \right)\ =\ 2\,gram$

A gas undergoes a process such that

$P \propto \dfrac{1}{T}$ . If the molar heat capacity for this process is

$24.93 \,J/mol \,K$ , then what is the degree of freedom of the molecules of the gas?

0%
$8$
0%
$4$
0%
$2$
0%
$6$

The degrees of freedom of a triatomic gas is? (consider moderate temperature)

0%
$6$
0%
$4$
0%
$2$
0%
$8$

Explanation

The general epression for degree of freedom is

$DOF=3N-n$

here, DOF means degree of freedom, N is number of particle, and n is the number of holonomic constraints.

for a triatomic molecule, the number of particle is 3 and since the separation between three atoms are fixed so, the number of constraints is 3.

hence,

$DOF=(3\times 3)-3$

$DOF=9-3$

$DOF=6$

Change in momentum of the gas molecules as they strike the walls
One mole of an ideal gas (mono-atomic) at temperature

$T_{0}$ expands slowly according to law

$P^{2} = cT$ (

$c$ is constant). If final temperature is

$2T_{0}$ , heat supplied to gas is

0%
$2RT_{0}$
0%
$\dfrac {3}{2}RT_{0}$
0%
$RT_{0}$
0%
$\dfrac {RT_{0}}{2}$

Which graph shows how the average kinetic energy of the particles varies with absolute temperature for an ideal gas?

The effect of temperature on Maxwell's speed distribution is correctly shown by

Select the incorrect relation. (Where symbols have their usual meanings)

0%
${ C }_{ p }=\dfrac { \gamma R }{ \gamma -1 }$
0%
${ C }_{ p }-{ C }_{ v }=R$
0%
$\Delta U=\dfrac { { P }_{ f }{ V }_{ f }-{ P }_{ i }{ V }_{ i } }{ 1-\gamma }$
0%
${ C }_{ v }=\dfrac { R }{ \gamma -1 }$

Explanation

Solution:- (C)

$\Delta{U} = \cfrac{{P}_{f}{V}_{f} - {P}_{i}{V}_{i}}{1 - \gamma}$

As we know that,

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

Whereas,

$n =$ No. of moles

${C}_{v} =$ Heat capacity at constant volume

$= \cfrac{R}{\gamma - 1}$

$\Delta{T} =$ Change in temperature

$= {T}_{f} - {T}_{i}$

$\therefore \Delta{U} = n \cfrac{R}{\gamma - 1} \left( {T}_{f} - {T}_{i} \right)$

$\Rightarrow \Delta{U} = \cfrac{nR{T}_{f} - nR{T}_{i}}{\gamma - 1} ..... \left( 1 \right)$

Now from ideal gas equation,

$PV = nRT$

For the same no. of moles of a gas,

$nR{T}_{f} = {P}_{f}{V}_{f}$

$nR{T}_{i} = {P}_{i}{V}_{i}$

Substituting these values in

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

$\Delta{U} = \cfrac{{P}_{f}{V}_{f} - {P}_{i}{V}_{i}}{\gamma - 1}$

Hence the given option

$\left( C \right) \Delta{U} = \cfrac{{P}_{f}{V}_{f} - {P}_{i}{V}_{i}}{1 - \gamma}$ is incorrect.

The ratio of adiabatic to isothermal elasticity of diatomic gas is:-

0%
$1.67$
0%
$1.4$
0%
$1.33$
0%
$1.27$

Explanation

The ratio of adiabatic to isothermal elasticity of diatomic gas is,

$\gamma =\dfrac{C_p}{C_v}$

$=\dfrac{C_v+R}{C_v}$

$=\dfrac{\frac{7R}{2}}{\frac{5R}{2}}$

$=\dfrac{7}{5}=1.4$

A vessel contains 28 gm of

$N-2$ and 32 gm of

$O_2$ at temperature T = 1800 K and pressure 2 atm pressure it

$N_2$ dissociates

$30%$ and

$O_2$ dissociates

$50%$ if temperature remains constant.

0%
2 atm
0%
1 atm
0%
2.8 atm
0%
1.4 atm

An ideal gas has an initial volume

$V$ and pressure

$P$ . In doubling its volume the minimum work done will be in (of the given processes):

0%
Isobaric process
0%
Isothermal process
0%
Adiabatic process
0%
Same in all given processes

The velocities of sound in an ideal gas at temperature

$T_1$ and

$T_2 k$ are found to be

$V_1$ and

$V_2$ respectively.If the r.m.s. velocities of the molecules of the same gas at the same temperatures

$T_1$ and

$T_2$ are

$v_1$ and

$v_2$ respectively then

0%
$v_2=v_1(\dfrac{V_1}{V_2})$
0%
$v_2=v_1(\dfrac{V_2}{V_1})$
0%
$v_2=v_1\sqrt{\dfrac{V_2}{V_1}}$
0%
$v_2=v_1\sqrt{\dfrac{V_1}{V_2}}$

For gas, if the ratio of specific heats at constants pressure

$P$ and constant volume

$V$ is

$\gamma$ , then the value of degree of freedom is:

0%
$\dfrac{\gamma +1}{\gamma -1}$
0%
$\dfrac{\gamma -1}{\gamma +1}$
0%
$\dfrac{1}{2}(\gamma-1)$
0%
$\dfrac{2}{\gamma-1}$

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%
$6.35$

Explanation

Let's assume we have

$1$ mole of triatomic gas

$\therefore 3Na$ is present

So,

$0.5$ moles=

$1.5 Na$ atoms

$1$ part of

$0.5$ moles remains untouched

Degree of dissociation=

$0.5 \times 6=3$

Degree of freedom for

$0.5Na= 1.5 \times 0.5=0.75$

Total=

$3+0.75=3.75$

At which of the following temperature would the molecules of a gas have twice the average kinetic energy they have at

${20}^{o}C$

0%
${40}^{o}C$
0%
${80}^{o}C$
0%
${313}^{o}C$
0%
${586}^{o}C$

Explanation

$\textbf{Step 1 - Average KE of gas molecule initial and final}$

Given,

$E_{2} = 2E_{1}$ and

$T_{1} = 20^{\circ}C$

$E_{1} = \dfrac {3}{2} KT_{1}$

$....(1)$

Similarly,

$E_{2} = \dfrac {3}{2} KT_{2}$

$....(2)$

$\textbf{Step 2 - Calculation of final temperature }$

Dividing equation (1) with (2)

$\dfrac {E_{1}}{E_{2}} = \dfrac {\dfrac {3}{2} KT_{1}}{\dfrac {3}{2} KT_{2}} = \dfrac {T_{1}}{T_{2}}$

$\dfrac {E_{1}}{2E_{1}} = \dfrac {(273 + 20)K}{T_{2}}$

$\Rightarrow T_{2} = 586\ K$

$^{\circ}C$

$T_{2} = 586 - 273$

$= 313^{\circ}C$

Correct option : C

The lowest pressure(the best Vaccum) that can be created in laboratory at 27 degree is

$10^{-11}$ mm of Hg. At this pressure, the number of ideal gass molecules per

$cm^{3}$ will be

0%
$3.22 \times 10 ^{12}$
0%
$1.61 \times 10 ^{12}$
0%
$3.21 \times 10 ^{6}$
0%
$3.22 \times 10 ^{5}$

The internal energy of an ideal gas increases during an isothermal process when the gas is

0%
Expanded by adding more molecules to it
0%
Expanded by adding more heat to it
0%
Expanded against zero pressure
0%
Compressed by doing work on it

Explanation

Internal energy of isotherma process

$v=\dfrac{f}{2} \times NKT$

$\Rightarrow v\propto NT$

T is constant (Isothermal process

$v \propto N$ i.e., Internal energy increases by increasing number of molecules(N).

One-gram mole of nitrogen occupies

$2 \times 10^4$ cc at a pressure of

$10^6 dynes/cm^2$ . The average energy of a nitrogen molecule (in erg) will be:

$(Avogadro's number = 6 \times 10^25 )$

0%
$14 \times 10^{-13}$
0%
$10 \times 10^{12}$
0%
$10^6$
0%
$2 \times 10^6$

The molecules of a given mass of gas have rms velocity

$200\ ms^{-1}$ at

$27^{\circ}C$ and

$10^5\ Nm^{-2}$ pressure. When the absolute temperature is doubled and the pressure is halved, the rms speed of the molecules of the same gas is:

0%
$200\ ms^{-1}$
0%
$400\ ms^{-1}$
0%
$200\sqrt{2}\ ms^{-1}$
0%
$400\sqrt{2}\ ms^{-1}$

In a thermodynamic system, working substance is ideal gas, its internal energy is in the form of

0%
kinetic energy
0%
kinetic and potential energy
0%
potential energy
0%
None of the above

$T_1$ is the temperature of oxygen enclosed in a cylinder. The temperature is increased to

$T_2$ and Maxwellan distribution curves for

$O_2$ at temperature

$T_1$ and

$T_2$ are plotted. If

$A_1$ and

$A_2$ are the areas under the curves and the speed axis, in both cases , then

0%
$A_1 > A_2$
0%
$A_1 < A_2$
0%
A_1 = A_2$$
0%
$A_1=\sqrt {A_2}$

Molar heat capacity of water in equilibrium with ice at constant pressure is

0%
Zero
0%
$\infty$
0%
40.45 kJ $K^{-1}mol^{-1}$
0%
75.48 $JK^{-1}mol^{-1}$

The speed of a longitudinal wave in a mixture containing 4 moles of He and 1 mole of Ne at 300 K will be

0%
930 m/s
0%
541 m/s
0%
498 m/s
0%
None of these

Explanation

The mixture contains

$n_1=4\ mol$ of

$He$ and

$n_2=4\ mole$ of

$Ne$ act.

$T=300\ K$ .

Forth

$He$ and

$Ne$ are monoatomic, so we take

$r$ mix

$=5/3\quad \left [r=\dfrac {cp}{cv}\right]$

$M'=\dfrac {M_1n_1+M_2n_2}{n_1+n_2}$

$=\dfrac {4\times 4+20\times 1}{4+1}=\dfrac {24}{5}=4.8\ g/mol$

$=4.8\times 10^{-3}\ kg/mol$

$\therefore \$ Speed of sound through mixture

$v=\sqrt {\dfrac {vRT}{M'}}=\sqrt {\dfrac {5}{3}\times \dfrac {8.3\times 300}{4.8\times 10^{-3}}}\sim 929.83\ m/s$

In a process the pressure of gas is inversely proportional to the square of volume If temperature of the gas is increased then work done by the gas

0%
Is positive
0%
Is negative
0%
Is zero
0%
May be positive

Hydrogen is a diatomic gas. Its molar specific heat at constant volume is very nearly

0%
$\frac { 3 R } { 2 }$
0%
$\frac { 5 R } { 2 }$
0%
$\frac { 7 R } { 2 }$
0%
(b) or (c) depending on the temperature.

In a thermodynamic system, working substance is ideal gas, its internal energy is in the form of

0%
Kinetic energy
0%
Kinetic and potential energy
0%
Potential energy
0%
None of the above

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 (2/3) rd the translational kinetic energy for each molecule.

Explanation

$(a,d)$ both are correct as translational kinetic energy and rotational kinetic energy both obey Maxwell's distribution independent of each other.
1380844_1208163_ans_f5e87849d26f49dd9f8e6a29234e3f01.png

1380844_1208163_ans_f5e87849d26f49dd9f8e6a29234e3f01.png

The total kinetic energy of

$8$ litres of helium molecules at

$5$ atmosphere pressure will be
(

$1$ atmosphere =

${ 1.013\times 10 }^{ 5 }$ pascal)

0%
$\text{607 J}$
0%
$\text{6078 J}$
0%
$\text{607 erg}$
0%
$\text{6078 erg}$

Explanation

$Given:$ The total K.E of 8 litres of helium molecules at 5 atmosphere pressure. (

$1 atmosphere=1.013\times10^{5} pascal)$

$Solution:$ We know that,

$\mathrm{E}_{\mathrm{k}}=\dfrac{3}{2} \mathrm{PV}$

$=\dfrac{3}{2} \times 5 \times 8 \times 1.013 \times 10^{5}=60.78 \times 10^{5}=6078 \mathrm{~J}$

$So,the$

$correct$

$option:B$

$2$ grams of mono atomic gas occupies a volume of

$2$ litres at a pressure of

$8.3 \times 10^5$ Pa and

$127^0C$ . Find the molecular weight of the gas.

0%
$2$ grams/mole
0%
$16$ grams/mole
0%
$4$ grams/mole
0%
$32$ grams/mole

Two moles of an ideal gas X occupying a volume V excert a pressure P. The same pressure is excert by one mole of another gas Y occupying a volume 2V. (if the molecular weight of Y is 16 times the molecular weight of X), find the ratio of the 'rms' speed of the molecular of X and Y.

0%
$2$
0%
$4$
0%
$\frac { 1 }{ 2 }$
0%
$\sqrt { 2 }$

The temperature of a diatomic gas is T. The total kinetic energy of the gas is given as

$E = { K }_{ 1 }$ where

${ K }_{ 1 }$ is constant. Find out the total number of molecules of the gas in the sample. (K = Boltzaman's constant)

0%
$3K_{ 1 }K$
0%
$5K_{ 1 }/2K$
0%
$2K_{ 1 }/5K$
0%
None of these

let A and B the two gases and given :

$\frac{{T}_{A}}{{M}_{A}}$ =

$\frac{{T}_{B}}{{M}_{B}}$ Where T is the temperature and M is molecular mass. If

${C}_{A}$ and

${C}_{B}$ are the r.m.s. speed, then the ratio

$\frac{{C}_{A}}{{C}_{B}}$ will be equal to:

0%
2
0%
4
0%
1
0%
0.5

If hydrogen gas is heated to a very high temperature, then the fraction of energy possessed by gas molecules correspond to rotational motion:-

0%
$\dfrac{3}{5}$
0%
$\dfrac{2}{7}$
0%
$\dfrac{3}{7}$
0%
$\dfrac{2}{5}$

A gas is filled in a cylinder. Its temperature is increased by

$20\%$ and volume decreases by

$20\%$ . If pressure remains constant, what percentage of mass of the gas leaks out?

0%
$33.33\%$
0%
$66.67\%$
0%
$20\%$
0%
$40\%$

The rms speed of

${ N }_{ 2 }$ molecular at STP (P=1 tm; T=

${ 0 }^{ 0 }C$ ) is.....
(the density of

${ N }_{ 2 }$ in these conditions is 1.25

${ kg/m }^{ 3 }$

0%
493 m/s
0%
390 m/s
0%
290 m/s
0%
590 m/s

Explanation

Given,

$P=1.013\times 10^5Pa$

$T=273.15 K$

$\rho =1.25kg/m^3$

The rms speed of

$N_2$ molecules is given by

$V_{rms}=\sqrt{\dfrac{3RT}{M}}=\sqrt{\dfrac{3P}{\rho}}$

$V_{rms}=\sqrt{\dfrac{3P}{\rho}}$

$V_{rms}=\sqrt{\dfrac{3\times 1.013\times 10^5}{1.25}}=493 m/s$

The correct option is A.

An ant is moving on a plane horizontal surface. The number of degrees of freedom of the ant will be

0%
1
0%
2
0%
3
0%
6

Explanation

$\textbf{Explanation:}$ A plane contains three axis i.e x, y and z axis.

The number of degrees of freedom of dynamical system is depends on the total numbers of coordinates or independent variables required to describe the position of the system.

When an ant moves on surface, it covers two axes, either xy, yz or zx.

So the degree of freedom of ant will be 2.

$\textbf{Option B is correct.}$

When

$2 gm$ of gas is introduced into an evacuated flask kept at

$25^oC$ the pressure is found to be one atmosphere. If

$3 gm$ of another gas added to the same flask the pressure becomes

$1.5$ atmospheres. The ratio of the molecular weights of there gases will be

0%
$1:3$
0%
$3:1$
0%
$1:2$
0%
$3:2$

$\gamma$ be the ration of specific heats of a perfect gas, the number of degree of freedom of a molecule of the gas is:

0%
$\dfrac{{25}}{2}\left( {\gamma - 1} \right)$
0%
$\dfrac{{3\gamma - 1}}{{2\gamma - 1}}$
0%
$\dfrac{2}{{\gamma - 1}}$
0%
$\dfrac{9}{2}(\gamma - 1)$

The degree of freedom of a diatomic gas at normal temperature is.

0%
$7$
0%
$6$
0%
$5$
0%
$4$

A real gas most closely approaches the behaviour of an ideal gas at :

0%
15 atm and 200 K
0%
1 atm and 273 K
0%
0.5 atm and 500 K
0%
15 atm and 500 K

When an ideal monoatomic gas is heated zt constant pressure , which of the following may be true

0%
$\dfrac {dU}{dQ} = \frac {3}{5}$
0%
$\dfrac {dW}{dQ} = \frac {2]}{5}$
0%
$\dfrac {dU}{dQ} = \frac {4}{5}$
0%
$dW + dU = dQ$

A gas mixture contains 8 moles of oxygen and 2 mole of argon at room temperature T. The total energy of the mixture is

0%
11 RT
0%
23 RT
0%
22 RT
0%
26 RT

The kinetic energy for 14 grams of nitrogen gas at

$127$ is nearly (mol. mass of nitrogen

$= 28$ and gas constant

$=8.31{ JK }_{ -1 }{ mol }_{ -1 }$ )

0%
$1.0J$
0%
$4.15J$
0%
$2493J$
0%
$3.3J$

CBSE Questions for Class 11 Engineering Physics Kinetic Theory Quiz 10 - 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 10 - MCQExams.com

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Practice Class 11 Engineering Physics Quiz Questions and Answers

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