MCQ Questions for Class 11 Engineering Physics Kinetic Theory Quiz 11

The maximum speed of the molecules of a gas in a vessel is

$400\ m/s$ . If half of the gas leaks out, at constant temperature, the

$rms$ speed of the remaining molecules will be-

0%
$800\ m/s$
0%
$400\sqrt2\ m/s$
0%
$400\ m/s$
0%
$200\ m/s$

Explanation

Given,

$v_{rms}=400m/s$

The RMS speed of the molecules will be

$v_{rms}=\sqrt{\dfrac{3RT}{M}}$ . . . . . .(1)

where

$R=$ universal gas constant

$T=$ temperature

$M=$ molecular weight

The RMS speed does not depend on the quantity of the gas. For a given gas and the temperature, it remains constant.

So, the RMS speed of the remaining molecules is

$400m/s$ .

The correct option is C.

Internal energy of

$n_{1}$ moles of hydrogen at temperature

$150\ K$ is equal to the internal energy of

$n_{2}$ moles of helium at temperature

$300\ K$ . The ratio of

$n_{1}/n_{2}$ is

0%
$3:5$
0%
$2:3$
0%
$6:5$
0%
$5:6$

Explanation

We have,

$u = \dfrac{5}{2} \times {n_1} \times R \times 150 = \dfrac{3}{2} \times {n_2} \times R \times 300$

$\Rightarrow \dfrac{{{n_1}}}{{{n_2}}} = \dfrac{{900}}{{750}} = \dfrac{6}{5}$

$\therefore$ Option

$C$ is correct answer.

Heat is added to an ideal gas, and the gas expands. In such a process the temperature

0%
must always increase
0%
will remain the same if the work done equals the heat added
0%
must always decrease
0%
will remains the same If change in internal energy equals the heat added.

Explanation

Heat is added to an ideal gas, and the gas expands must always increases.

Hence Option

$A$ is correct.

If number of molecules of

${ H }_{ 2 }$ are doubles than that of

${ O }_{ 2 }$ , then ratio of kinetic energy of hydrogen to that of oxygen at 300 K is

0%
1.1
0%
1.2
0%
2.1
0%
1.16

On increasing temperature of the reacting system by

$10$ degrees the rate of reaction almost doubles. The most appropriate reason for this is

0%
collision frequency increases
0%
activation energy decreases by increases in temperatuer
0%
the fraction of molecules having energy equal to threshold energy or more increase
0%
the value of threshold energy decreases

Two spherical vessel of equal volumes, are connected by a narrow tube. The apparatus contains an ideal gas at one atmosphere and 300K. Now if one vessel is immersed in a bath of constant temperature 600 K and the other in a bath of constant temperature 300 K. Then the common pressure will be

0%
$1 atm$
0%
$4/5 atm$
0%
$4/3 atm$
0%
$3/4 atm$

Select the correct statement

0%
The average KE of an ideal gas in calories per mole is approximately equal to its absolute temperature
0%
The average KE of an ideal gas in calories per mole is approximately equal to two times its absolute temperature
0%
The average KE of an ideal gas in calories per mole is approximately equal to three times its absolute temperature
0%
The average KE of an ideal gas in calories per mole is approximately equal to 1.5 times its absolute temperature

n moles of an ideal monoatomic gas undergoes an isothermal expansion at temperature T during which its volume becomes 4 times. The work done on the gas and change in internal energy of the gas respectively is

0%
n RT Ln 4,0
0%
- n RT Ln 4,0
0%
n RT Ln 4 $\frac { 3 n R T } { 2 }$
0%
- n RT Ln 4, $\frac { 3 n R T } { 2 }$

Explanation

$\begin{array}{l} w=nRT\, \, \, \ln { \left( { \frac { { 4v } }{ v } } \right) } \\ =nRT\, \, \ln { 4 } \, \, \, \, \, \, \& \, \, \, \Delta u=0 \end{array}$

$\therefore$ Option

$A$ is correct.

Air is filled at

$60^0$ C in a vessel of open mouth. The vessel is heated to a temperature T so that

$1/4^{th}$ part of air escapes. The value of T is :

0%
$80^0$ C
0%
$444^0$ C
0%
$333^0$ C
0%
$171^0$ C

The magnetic monment of a diamagnetic atom is

0%
Much greater than one
0%
1
0%
Between zero and one
0%
Equal to zero

$'n'$ moles of an ideal gas undergoes a process

$A\rightarrow B$ as shown in the figure. The maximum temperature of the gas during the process will be :

0%
$\cfrac{3P_{0}V_{0}}{2nR}$
0%
$\cfrac{9P_{0}V_{0}}{2nR}$
0%
$\cfrac{9P_{0}V_{0}}{nR}$
0%
$\cfrac{9P_{0}V_{0}}{4nR}$

Explanation

$\begin{array}{l} y-{ y_{ 1 } }=\dfrac { { { y_{ 2 } }-{ y_{ 1 } } } }{ { { x_{ 2 } }-{ x_{ 1 } } } } \left( { x-{ x_{ 1 } } } \right) \\ P-{ P_{ 0 } }=\dfrac { { 2{ p_{ 0 } }-{ p_{ 0 } } } }{ { { v_{ 0 } }2{ v_{ 0 } } } } \left( { V-2{ V_{ 0 } } } \right) \\ =\dfrac { { -{ P_{ 0 } } } }{ { { V_{ 0 } } } } \left( { V-2{ V_{ 0 } } } \right) \\ P=\dfrac { { -{ P_{ 0 } } } }{ { { V_{ 0 } } } } V+3{ p_{ 0 } } \\ PV=\dfrac { { -{ P_{ 0 } } } }{ { { V_{ 0 } } } } V+3{ p_{ 0 } }V \\ nRT=\dfrac { { -{ P_{ 0 } } } }{ { { V_{ 0 } } } } V+3{ p_{ 0 } }V \\ T=\dfrac { 1 }{ { nR } } \left( { \dfrac { { -{ P_{ 0 } } } }{ { { V_{ 0 } } } } { V^{ 2 } }+3{ p_{ 0 } }V } \right) \\ \dfrac { { dT } }{ { dV } } =0\, \, \left( { for\, \, \max \, \, \, temperature } \right) \\ \dfrac { { -{ P_{ 0 } } } }{ { { V_{ 0 } } } } 2V+3{ p_{ 0 } }=0 \\ \dfrac { { -{ P_{ 0 } } } }{ { { V_{ 0 } } } } 2V=-3{ p_{ 0 } } \\ V=\dfrac { 3 }{ 2 } { V_{ 0 } }\, \, \, \, \left( { condtion\, \, for\, \, \max \, \, temperature } \right) \\ { T_{ \max } }=\dfrac { 1 }{ { nR } } \left( { \dfrac { { -{ P_{ 0 } } } }{ { { V_{ 0 } } } } \times \dfrac { 9 }{ 4 } V_{ 0 }^{ 2 }+3{ p_{ 0 } }\times \dfrac { 3 }{ 2 } { V_{ 0 } } } \right) \\ { T_{ \max } }=\dfrac { 1 }{ { nR } } \left( { -\dfrac { 9 }{ 4 } { P_{ 0 } }{ V_{ 0 } }+\dfrac { 9 }{ 2 } { P_{ 0 } }{ V_{ 0 } } } \right) \\ =\dfrac { 9 }{ 4 } \dfrac { { { P_{ 0 } }{ V_{ 0 } } } }{ { nR } } \end{array}$

Find the approximate number of molecules contained in a vessel of volume

$7$ litres at

$0^oC$ at

$1.3\times 10^5$ pascals:

0%
$2.4\times 10^{23}$
0%
$1.5\times 10^{24}$
0%
$6\times 10^{23}$
0%
$4.8\times 10^{23}$

Where temperature of a gas, contained in closed vessel is increased by

$5C$ , its pressure increases by

$1%$ .The original temperature of the gas was approx i - mately:-

0%
$500C$
0%
$273C$
0%
$227C$
0%
$50C$

A rigid container of negligible heat capacity contains one mole of an ideal gas. The temperature of the gas increase by

$1^{o}C$ if

$3 \cdot 0\ cal$ of heat is added to it. The gas may be

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

Explanation

Option A and B are correct.

Given,

n=1, dT=

$1^oC$ , dQ=3cal=12.54joule

We know that,

$dQ=nC_vdT$

$12.54=1 \times C_v \times1$

$C_v=12.54$

We also know

$C_v=\dfrac{3}{2}R$

$C_v=1.5\times8.314$ (R=8.314)

$C_v=12.54$

Since both have equal values gas is monoatomic,

Since Helium and argon are monoatomic gases.

Option A and B are correct.

An ant is walking on the horizontal surface. The number of degree of freedom of ant will be

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

The mass of hydrogen molecules is

$3.32 \times 10 ^ { - 24 }$ gm. If

$10 ^ { 23 } \mathrm { H } _ { 2 }$ molecules strike

$2$ sq. cm are per second with velocity of

$10 ^ { 5 }$ cm/sec at an angle of

$45 ^ { \circ }$ to the normal to wall, then the exerted pressure will be

0%
$2.35 \mathrm { N } / \mathrm { m } ^ { 2 }$
0%
$23.5 \mathrm { N } / \mathrm { m } ^ { 2 }$
0%
$235 \mathrm { N } / \mathrm { m } ^ { 2 }$
0%
$2350 \mathrm { N } / \mathrm { m } ^ { 2 }$

Explanation

$\begin{array}{l} \Delta P=\left( { 2mv\cos { { 45 }^{ o } } } \right) \times { 10^{ 23 } } \\ =2\times 3.32\times { 10^{ -24 } }\times { 10^{ -3 } }\times \frac { { 10 } }{ { 100 } } \times \frac { 1 }{ { \sqrt { 2 } } } \times 10 \\ \Rightarrow F=\frac { { 2\times 3.32 } }{ { \sqrt { 2 } } } \times { 10^{ -1 } } \\ \therefore pressure=\frac { { 2\times 3.32\times { { 10 }^{ -1 } } } }{ { \sqrt { 2 } \times 2\times { { 10 }^{ -4 } } } } \\ =\frac { { 2\times 3.32 } }{ { 2\sqrt { 2 } } } \times { 10^{ 3 } } \\ =\frac { { 3.32 } }{ { \sqrt { 2 } } } \times { 10^{ 3 } }\, N/{ m^{ 2 } } \\ \approx 2350\, N/{ m^{ 2 } } \\ Hence, \\ option\, \, D\, \, is\, \, correct\, \, answer. \end{array}$

Which of the following is incorrect relation fot a perfect gas?

0%
${ \quad \left( \frac { \eth U }{ \eth V } \right) }_{ T }=0$
0%
${ \quad \left( \frac { \eth H }{ \eth P } \right) }_{ T }=0$
0%
${ \quad \left( \frac { \eth T }{ \eth P } \right) }_{ H }=0$
0%
${ \quad \left( \frac { \eth U }{ \eth T } \right) }_{ V }=0$

Explanation

$(A)\left(\dfrac{\partial U}{\partial V}\right)_T=0$

because perfect gas equation is

$PV=nRT.$

$\left(\dfrac{\partial H}{\partial P}\right)_T=\left(\dfrac{\partial T}{\partial P}\right)_H=\left(\dfrac{\partial U}{\partial b}\right)_V$

If the law of equipartition of energy is applied to solid elements like silver, the heat capacity of one gram atom of the solid is

0%
$\dfrac5 2 R$
0%
$\dfrac32 R$
0%
$R$
0%
$3 R$

When a monoatomic gas expands at consatnt pressure, the percentage of heat supplied that increases temperature of the gas and in doing external work in expansion at constant pressure is

0%
$100%,0$
0%
$60%,40%$
0%
$40%,60%$
0%
$75%,25%$

An ideal gas is expanding such that

$PT = constant$ . The coefficient of volume expansion of the gas is

0%
$\dfrac {1}{T}$
0%
$\dfrac {2}{T}$
0%
$\dfrac {3}{T}$
0%
$\dfrac {4}{T}$

The Kinetic energy of translation of

$20$ gram of oxygen at

${47^0}C$ is [ Molecular weight of

${O_2} = 32$ gram/ mol and

$R = 8.3J/mol/K$ ]

0%
2490 ergs
0%
2490 J
0%
249 J
0%
960 J

Two kg of a monoatomic gas is at a pressure of

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

$8 kg /m^3$ . What is the order of energy of the gas due to its thermal motion ?

0%
$10^3 J$
0%
$10^5 J$
0%
$10^6 J$
0%
$10^4 J$

Explanation

Thermal energy of N molecule

$= N \left(\dfrac{3}{2} kT \right)$

$= \dfrac{N}{N_A} \dfrac{3}{2} RT$

$= \dfrac{3}{2} (nRT)$

$= \dfrac{3}{2} PV$

$= \dfrac{3}{2} P \left(\dfrac{m}{8} \right)$

$= \dfrac{3}{2} \times 4 \times 10^4 \times \dfrac{2}{8}$

$= 1.5 \times 10^4$
order will

$10^4$ .

The kinetic energy of 1 gram mole of a gas at normal temperature and pressure is ( R = 8.31 J/mole- K)

0%
0.56 $\times { 10 }^{ 4 }J$
0%
1.3 $\times { 10 }^{ 2 }J$
0%
2.7 $\times { 10 }^{ 2 }J$
0%
3.4 $\times { 10 }^{ 3 }J$

Explanation

Given,

$R=8.31J/mol K$

$T=273K$

Kinetic energy,

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

$K=\dfrac{3}{2}\times 8.31\times 273=3.4\times 10^3 J$

The correct option is D.

One

$kg$ of diatomic gas is at a pressure of

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

$4\ kg/m^{3}$ . What is the energy of the gas due to its thermal motion?

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

Explanation

Thermal energy corresponds to internal energy.

Mass

$=1$ kg

Density

$=8$ kgper

${m}^{3}$

Volume

$=\dfrac{Mass}{Density}=\dfrac{1}{8}{m}^{3}$

Pressure

$=8\times{10}^{4}$ Nper sq.m

$\therefore$ Internal Energy

$=\dfrac{5}{2}PV=5\times {10}^{4}$ J

For ideal monoatomic gas,the universal gas constant R is n times molar heat capacity at constant pressure

$C_p$ Here n is :-

0%
0.67
0%
1.4
0%
0.4
0%
1.67

A mixture Of

$n _ { 2 }$ moles of mono atomic gas and

$n _ { 2 }$ moles of diatomic gas has

$\frac { C _ { p } } { C _ { V } } = y = 1.5$

0%
$n _ { 1 } = n _ { 2 }$
0%
$2 n _ { 1 } = n _ { 2 }$
0%
$n _ { 1 } = 2 n _ { 2 }$
0%
$2 n _ { 1 } = 3 n _ { 2 }$

Explanation

$\begin{array}{l} As\, { y_{ mix } }=\dfrac { { { C_{ { p_{ mix } } } } } }{ { { C_{ { v_{ mix } } } } } } where\, { C_{ { p_{ mix } } } }=\dfrac { { { n_{ 1 } }{ C_{ { p_{ 1 } } } }+{ n_{ 2 } }{ C_{ { p_{ 2 } } } } } }{ { { n_{ 1 } }+{ n_{ 2 } } } } \\ and\, \, { C_{ { v_{ mix } } } }=\dfrac { { { n_{ 1 } }{ C_{ { v_{ 1 } } } }+{ n_{ 2 } }{ C_{ { v_{ 2 } } } } } }{ { { n_{ 1 } }+{ n_{ 2 } } } } \\ So,\, \, { y_{ mix } }=\dfrac { { { n_{ 1 } }{ C_{ { p_{ 1 } } } }+{ n_{ 2 } }{ C_{ { p_{ 2 } } } } } }{ { { n_{ 1 } }+{ n_{ 2 } } } } \\ Given\, ,\, for\, monoatomic\, { C_{ p } },\dfrac { 5 }{ 2 } R\, and\, { C_{ { v_{ 1 } } } }=\dfrac { 3 }{ 2 } R \\ For\, diatomic\, { C_{ { p_{ 2 } } } }=\dfrac { { 7R } }{ 2 } \, and\, { C_{ { v_{ 2 } } } }=\dfrac { 5 }{ 2 } R \\ { y_{ mix } }=\dfrac { { { n_{ 1 } }\times \dfrac { 5 }{ 2 } R+{ n_{ 2 } }\times \dfrac { 7 }{ 2 } R } }{ { { n_{ 1 } }\times \dfrac { 3 }{ 2 } R+{ n_{ 2 } }\times \dfrac { 5 }{ 2 } R } } =\dfrac { { 5{ n_{ 1 } }+7{ n_{ 2 } } } }{ { 3{ n_{ 1 } }+5{ n_{ 2 } } } } =\dfrac { 3 }{ 2 } \\ 10{ n_{ 1 } }+14{ n_{ 2 } }=9{ n_{ 1 } }+15{ n_{ 2 } } \\ { n_{ 1 } }={ n_{ 2 } } \\ Hence, \\ option\, \, A\, \, is\, correct\, \, answer. \end{array}$

An ideal gas at pressure P is adiabatically compressed so that its density becomes n times the initial value. The final pressure of the gas will be

$( \gamma= C_p/C_v)$

0%
$n^\gamma P$
0%
$n^{-\gamma} P$
0%
$n^{\gamma-1} P$
0%
$n^{1-\gamma} P$

The gas mixture constists of

$3$ moles of oxygen and

$5$ moles of argon at temperature

$T$ . Considering only translational and rotational modes, the total internal energy of the system is:

0%
$12\ RT$
0%
$20\ RT$
0%
$15\ RT$
0%
$4\ RT$

Explanation

$U=\dfrac{f_1}{2}n_1RT+\dfrac{f_2}{2}n_2RT$

$=\dfrac{5}{2}(3RT)+\dfrac{3}{2}\times 5RT$

$U=15RT$

The volume of an ideal gas is 1 litre and its pressure is equal to 72 cm of mercury column. The volume of gas is made

$900 cm^3$ by compressing it isothermally. The stress of the gas will be

0%
8 cm (mercury)
0%
7 cm (mercury)
0%
6 cm (mercury)
0%
4 cm (mercury)

Two moles of helium gas are taken along the path ABCD (as shown in figure). The work done by the gas is

0%
$2000R(1 + ln \frac{4}{3})$
0%
$500R(3+ln4)$
0%
$500R(2+ln\frac{16}{9})$
0%
$1000R(1+ln\frac{16}{9})$

What is the mass of :

0%
$0.2$ mole of oxygen atoms?
0%
$0.5$ mole of water molecules?
0%
Both a and b
0%
None of these

An open container is placed in atmosphere. During a time interval, temperature of atmosphere increases and then decreases. The internal energy of gas in container

0%
First increases then decreases because internal energy depends upon temperature
0%
First decreases then increases
0%
Remains constant
0%
Is inversely proportional to temperature

Half mole of an ideal monoatomic gas is heated at constant pressure of 1 atm from

${ 20 }^{ 0 }C$ to

${ 90 }^{ 0 }C$ . Work done by gas is close to : ( gas constant R=8.31 j / mol -k )

0%
146 j
0%
581 j
0%
73 j
0%
291 j

The mean free path of nitrogen molecules at a pressure of 1.0 atm and temperature

$0^0 C$ is

$0.8 \times 10^{-7} m$ . If the number of density of molecules is

$2.7 \times 10^{25}$ per

$m^3$ , then the molecular diameter is

0%
$0.32 nm$
0%
$3.2 A^0$
0%
$3.2 \mu m$
0%
$2.3 mm$

One mole of an ideal gas expands against a constant external pressure of 1 atm from a volume of

$10d{ m }^{ 3 }$ to a volume of

$30d{ m }^{ 3 }$ . What would be the work done in joules?

0%
$-2026 J$
0%
$+2026 J$
0%
$-1947 J$
0%
$1648 J$

Explanation

$\begin{array}{l} Work\, done\, for\, isothermal\, ansion,W=-\int _{ { V_{ 1 } } }^{ { V_{ 2 } } }{ pdV } \\ { { Sin } }ce,\, \rho \, is\, cons\tan t, \\ W=-\int _{ { V_{ 1 } } }^{ { V_{ 2 } } }{ dV } \\ W=-\rho \left( { { V_{ 2 } }-{ V_{ 1 } } } \right) \\ W=-1\, atm\left( { 30-10 } \right) L \\ =-20\, atm\, L \\ Now,\, 0.08206\, L\, atm=8.314J\left( { As\, we\, know } \right) \\ -20\, atm\, L=\dfrac { { 8.314 } }{ { 0.08206 } } \times -20J \\ =-2026.32J \end{array}$

Hence,

option

$(A)$ is correct answer.

The temperature at which the rms of a hydrogen molecule is equal to that of an oxygen molecule at

$47^{\circ}C$ is

0%
-253 K
0%
-73 K
0%
3 K
0%
20 K

A sample of gas is at

$0^{\circ}C$ . To what temperature must it be raised in order to double the rms speed of its molecules?

0%
$102^{\circ}C$
0%
$273^{\circ}C$
0%
$819^{\circ}C$
0%
$1092^{\circ}C$

Which of the following statements about kinetic theory of gases is wrong

0%
The molecules of a gas are in continuous random motion
0%
The molecules continuously undergo inelestic collisions
0%
The molecules do not interact with each other excepted during collisions
0%
The collisions among the molecules are of short duration

N moles of a monoatomic gas is carried round the reversible rectangular cycle ABCDA as shown in the diagram. The temperature at

$A$ is

$T _ { 0 }$ . The thermodynamic efficiency of the cycle is

0%
$15 \%$
0%
$50 \%$
0%
$20 \%$
0%
$25 \%$

Explanation

$\begin{array}{l} At\, \, A\, \, { P_{ 0 } }{ V_{ 0 } }=nR{ T_{ 0 } }\to \left( i \right) \\ At\, \, B\, \, 2{ P_{ 0 } }{ V_{ 0 } }=nR{ T_{ 1 } }\to \left( { ii } \right) \end{array}$

Efficiency of thermodynamic cycle

$\eta = 1 - \frac{{{T_B}}}{{{T_A}}}$

$By\left( i \right)\,\,and\,\,\left( {ii} \right)$

$\dfrac{{{T_B}}}{{{T_A}}} = \dfrac{1}{2}$

$\eta = 1 - \frac{1}{2} = \dfrac{1}{2} = 50\,\,\%$

Option B.

Find the amount of work done to increase the temperature of 1 mol of an ideal gas by

$30 ^ { \circ } { C }$ if it is expanding under the condition

$V \propto T ^ { 2 / 3 }$ .

0%
$166.2$ ${ J }$
0%
$136.2$ ${ J }$
0%
$126.2$ ${ J }$
0%
None of these

The pressure in the tyre of a car is four times the atmospheric pressure at

$300\, K$ . If this tyre suddenly bursts, its new temperature will be

$(\gamma = 1.4)$

0%
$300(4)^{1.4/0.4}$
0%
$300\,\left(\dfrac{1}{4}\right)^{-0.4/1.4}$
0%
$300(2)^{-0.4/1.4}$
0%
$300(4)^{-0.4/1.4}$

Average kinetic energy per mole if an monoatomic ideal gas at

0%
$5.8\times { 10 }^{ -2 }\quad J$
0%
$9.5\times { 10 }^{ -2 }\quad J$
0%
$5.6\times { 10 }^{ -2 }\quad J$
0%
$12.47 J$

The specific heat of a monatomic gas at constant volume is

$0.075 kcal$

$k g^{-1} K^{-1} .$ Its atomic weight will be

0%
$10$
0%
$30$
0%
$40$
0%
$90$

A vessel contains 3.20 g of oxygen and

$2.80 g$ of nitrogen. The vessel is moving with a velocity

$100 \mathrm { m } / \mathrm { s }$ . The vessel is suddenly stopped. Find the change in temperature -

0%
$7.2 ^ { \circ } \mathrm { C }$
0%
$6.0 ^ { \circ } \mathrm { C }$
0%
$8.3 ^ { \circ } \mathrm { C }$
0%
$6.5 ^ { \circ } \mathrm { C }$

The mass of litre of dry air at N.T.P is 1.293 g . Find the mass of 3 litres of air at

$117^oC$ and a pressure of 4 atmospheres _____.

0%
10.86 g
0%
5.6 g
0%
6.4 g
0%
7.2 g

The specific heat of a substance is

$0.09 cal/gm^oC$ . If the temperature is measured on Pahrenhelt scale the value of its specific heat in

$cal/gm/^oF$ is _________.

0%
0.05
0%
0.9
0%
0.005
0%
0.5

Heat capacity at constant temperature and constant pressure for

${ H }_{ 2 }$ is :-

0%
5 cal ${ moI }^{ -1 }{ K }^{ -1 }$
0%
7 cal ${ moI }^{ -1 }{ K }^{ -1 }$
0%
8 cal ${ moI }^{ -1 }{ K }^{ -1 }$
0%
$\infty$

If the temperature of the gas becomes three times the initial absolute temperature, speed of the gas molecules

0%
Becomes $\dfrac{1}{ 3}$ times the initial ms speed
0%
Becomes $\dfrac{1}{\sqrt { 3 }}$ times the initial rms speed
0%
Becomes $\sqrt { 3 }$ times times the initial ms speed
0%
Becomes 3 times the initial rms speed

The temperature of a gas is

$-{ 68 }^{ \circ }C$ At what temperature will the average kinetic energy of its molecules be twice that of

$-{ 68 }^{ \circ }C$ ?

0%
$137^{ \circ }C$
0%
$127^{ \circ }C$
0%
$100^{ \circ }C$
0%
$105^{ \circ }C$

An ideal gas is expanding such that

$P T ^ { 2 } =$ constant. The coefficient of volume expansion of the gas is

0%
$\dfrac{1}{T}$
0%
$\dfrac{2}{T}$
0%
$\dfrac{3}{T}$
0%
$\dfrac{4}{T}$

Explanation

Correct Answer: Option C

Hint: Using ideal gas equation rearrange the given equation and then differentiate.

Where P - Pressure, T - Temperature

Consider the ideal gas equation and substitute the value of pressure in the given equation.

Explanation of Correct Option:

Step 1: Consider the formulas mentioned

Expression of is given by,

Volume expansion can be expressed as:

$V={{V}_{0}}(1+\gamma t)$

Where V is the volume at temperature t

${{V}_{0}}$ Is the volume at temperature t = 0

$\gamma$ is the coefficient of volume expansion and t is temperature.

Step 2: Consider the ideal gas expansion

Ideal gas is expansion is given as

$P{{T}^{2}}=\text{constant}$ ….. (1)

Step 3: Consider the ideal gas equation,

$PV=nRT$

$\dfrac{PV}{T}=\text{constant}$

$P=\dfrac{\text{constant}\times T}{V}......\left( 2 \right)$

Coefficient of volume expansion can be calculated by the values of gas denoted by $\gamma$

Step 4: Coefficient of volume expansion of gas is given by

$V={{V}_{0}}(1+\gamma t).......\left( 3 \right)$

Substitute the equation (2) in equation (1),

$\dfrac{\text{constant}\times T}{V}\times {{T}^{2}}=\text{constant}$

$\dfrac{{{T}^{3}}}{V}=\text{constant=k}$

$V=\dfrac{1}{k}{{T}^{3}}={{k}^{1}}{{T}^{3}}.....\left( 4 \right)\text{ }\left( \because {{k}^{1}}=\dfrac{1}{k} \right)$

Step 5: Differentiate equation (3) with respect to t

On differentiating with respect to t

$\dfrac{dV}{dt}={{V}_{0}}(\gamma .1)$

$\dfrac{dV}{dt}={{V}_{0}}\gamma .......\left( 5 \right)$

Dedifferentiate equation (4) with respect to T,

$\dfrac{dV}{dT}={{k}^{1}}3{{T}^{2}}$

Put values of ${{k}^{1}}$ from equation (4)

$\dfrac{dV}{dT}=\dfrac{V}{{{T}^{3}}}\times 3{{T}^{2}}$

$\dfrac{dV}{dT}=\dfrac{3V}{T}$

$\therefore \dfrac{dV}{V}=3\dfrac{dT}{T}$

$dV=\dfrac{3}{T}VdT......(6)$

Volume expansion is given by,

$\Delta V=\gamma {{V}_{0}}\Delta T......\left( 7 \right)$

Step 6: Compare the equations (6) and (7)

$dV=\Delta V$

$\dfrac{3}{T}VdT = \gamma {{V}_{0}}\Delta T......\left( 7 \right)$

$\gamma =\dfrac{3}{T}$

Hence the coefficient of volume expansion of the gas is $\gamma =\dfrac{3}{T}$ .

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

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Practice Class 11 Engineering Physics Quiz Questions and Answers

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