The heat capacity of a certain amount of a particular gas at constant pressure is greater than that at constant volume by 29.1 J/K. Match the items given in Column I with the items given in Column II. List 1 List 2 If the gas is monatomic, heat capacity at constant volume 131 J/K If the gas monatomic, heat capacity at constant pressure 43.7 J/K If the gas is rigid diatomic, heat capacity at constant pressure 72.7 J/K If the gas is vibrating diatomic, heat capacity at constant pressure 102 J/K

A$$\rightarrow$$2, B$$\rightarrow$$3, C$$\rightarrow$$4, D$$\rightarrow$$1

B$$\rightarrow$$1, C$$\rightarrow$$3, D$$\rightarrow$$4, A$$\rightarrow$$1

C$$\rightarrow$$1, D$$\rightarrow$$3, A$$\rightarrow$$4, B$$\rightarrow$$1

D$$\rightarrow$$1, A$$\rightarrow$$3, B$$\rightarrow$$4, C$$\rightarrow$$1

MCQ Questions for Class 11 Engineering Physics Kinetic Theory Quiz 14

A closed hollow insulated cylinder is filled with gas at 0 $^{0}$ C and also contains an insulated piston of negligible weight and negligible thickness at the the middle point. The gas at one side of the piston is heated to 100 $^{0}$ C . If the piston moves 5cm, the length of the hollow cylinder is

0%
13.65 cm
0%
27.3 cm
0%
64.6 cm
0%
54.6 cm

Explanation

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

For section with constant temperature,

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

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

Similarly for section whose temperature increased,

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

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

$\implies L=64.6cm$

An ideal monoatomic gas

$(C_{v}=1.5\ R)$ initially at

$298$ K and

$1.013$ atm expands adiabatically irreversibly until it is in equilibrium with a constant external pressure of

$0.1013$ atm. The final temperature (in Kelvin) of the gas is :

0%
$190.7$
0%
$119.7$
0%
$278$
0%
$273$

Explanation

We know the relation,

$C_{v}=\dfrac{R}{\gamma-1}$ , where

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

$C_p - C_v = R$

$PV^{\gamma}=$ constant for adiabatic process (reversible)

$W=C_{v}(T_{2}-T_{1})=-RP_{ext}\left ( \dfrac{T_{2}}{P_{2}}-\dfrac{T_{1}}{P_{1}} \right )$

Substituting the values, we get-

$1.5\times R\times (-298+T_{2})=-R \times (0.1013)\left ( \dfrac{T_{2}}{0.1013}-\dfrac{298}{1.013} \right )$

$\implies$

$1.5\times R\times (T_{2}-298)=-R \times(T_{2}-29.8)$

$\implies$

$1.5\times T_{2}-1.5\times 298=-T_{2}+29.8$

$\implies$

$2.5\times T_{2}=1.6\times 298$

$\implies$

$T_{2}=\dfrac{1.6\times 298}{2.5}$

$=190.72\ K$

Option A is correct.

How many degrees of freedom have the gas molecules if at S.T.P the gas density is 1.3 kg/m

$^{3}$ and the velocity of sound in the gas is 330 m/sec?

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

Explanation

Velocity of sound in the gas is given by

$v=\sqrt \frac {\gamma P}{\rho}$ where
P is pressure,

$\rho$ is the density of the gas and

$\gamma=\frac {C_p}{C_v}$ is the ratio of specific heats at constant pressure to constant volume.

$\therefore \gamma = \dfrac {330^2 \times 1.3}{1.013 \times 10^5} =1.4$

But,

$\gamma = \dfrac{2+f}{f}$ where f is the no of degrees of freedom.
Solving

$1.4= \dfrac{2}{f} +1$

$\therefore \dfrac{2}{f} = 0.4$

$\therefore f=5$

A motor draws some gas from an adiabatic container of volume 2 liters having a monatomic gas at a pressure 4 atm. After drawing the gas the pressure in the container reduces to 1 atm. If the motor converts 10% of the energy contained in the drawn gas and the output of the motor is 10p joule, then p is
( Use

$1atm=10^5 pascal)$

0%
9
0%
4
0%
8
0%
6

Explanation

We have internal energy of a gas in a container,

$U = \dfrac{f}{2} n R T$

$= \dfrac {f}{2} PV$ (as

$PV = nRT$ )

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

$f = 3$ for a monoatomic gas)

The initial internal energy is given by

$U_i = \dfrac {3}{2} \times 4 \times 2 = 12 atm-litre$

The final internal energy is given by

$U_f = \dfrac {3}{2} \times 1 \times 2 = 3 atm-litre$

Total energy of the gas drawn

$= 9 atm-litre = 911.7 J$ (

$1 atm-litre = 101.3J$ )

Output of the motor = 10 % of 911.7 = 91.2 J ~ 90 J =10 p

Hence

$p = 9$

Which of the following will have maximum total kinetic energy at temperature 300 K ?

0%
$1 kg, H_{2}$
0%
$1 kg , He$
0%
$\frac{1}{2}kg H_{2}+\frac{1}{2}kg He$
0%
$\frac{1}{4}kg H_{2}+\frac{3}{4}kg He$

Explanation

We have total Kinetic energy,

$U = \frac {f}{2} nRT$
The total kinetic energy in the case of 1 kg of

$H_2$ is the greatest as the degree of freedom is 5 and number of moles is the highest as it contains 1 kg mass.

A gas mixture consists of

$2$ moles of oxygen and

$4$ moles of argon at temperature

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

0%
$4 RT$
0%
$5 RT$
0%
$15 RT$
0%
$11 RT$

Explanation

The internal energy of

$n$ moles of a gas is

$u = \dfrac {1}{2} nFRT$
where

$F =$ number of degrees of freedom.
The internal energy of

$2\ moles$ of oxygen at temperature

$T$ is

$u_{1} = \dfrac {1}{2}\times 2\times 5RT = 5RT (F = 5$ for oxygen molecule)
Total internal energy of

$4$ moles of argon at temperature

$T$ is

$= u_{2} = \dfrac {1}{2}\times 4\times 3RT = 6RT$
Total internal energy

$= u_{1} + u_{2} = 11\ RT$ .

2 moles of an ideal gas A (

${C}_{P}$

$= 4R$ ) and 4 moles of an ideal gas B (

${C}_{V}$

$= \dfrac {3R}{2}$ ) are taken together in a container and allowed to expand reversibly and adiabatically from 49L to 64L, starting from an initial temperature of

${47}^{0}$ C. The final temperature of the gas is :

0%
$4{}^{0}C$
0%
$5{}^{0}C$
0%
$7{}^{0}C$
0%
$8{}^{0}C$

Explanation

${\gamma}_{min} = \dfrac {{n}_{A}{C}_{{P}_{A}} + {n}_{B}{C}_{{P}_{B}}}{{n}_{A}{C}_{{V}_{A}} + {n}_{B}{C}_{{V}_{B}}} = \dfrac {2 \times 4R + 4 \times \dfrac {5R}{2}}{2 \times 3R + 4 \times \dfrac{3R}{2}}$

$= \dfrac {18R}{12R} = 1.5$

Now,

${T}_{1}{V}_{1}^{\gamma - 1} = {T}_{2}{V}_{2}^{\gamma - 1}$

$320\times{(49)}^{0.5} = {T}_{2}\times {(64)}^{0.5}$

${T}_{2} = \dfrac {320 \times 7}{8} =280 K = {7}^{0}C$

The ratio of translational and rotational kinetic energies at 100 K temperature is 3 :Then the internal energy of one mole gas at that temperature is

$[R = 8.3 J/mol-K]$

0%
$1175J$
0%
$1037.5 J$
0%
$2075 J$
0%
$4150J$

Explanation

According to law of equipartition of energy, energies
equally distributed among its degree of freedom,
Let translational and rotational degree of freedom be

$f_1$ and

$f_2$ .

Therefore

$\dfrac{K_T}{K_R}=\dfrac{3}{2}$
and

$K_T+K_R=U$

Hence the ratio of translational to rotational degrees of freedom is

$3:2$ . Since translational degrees of freedom is 3, the rotational degrees of freedom must be 2.
Internal energy

$U=1\times (f_1+f_2)\times \dfrac{1}{2}RT$

$=\dfrac{5}{2}RT=2075J$

One mole of an ideal gas

$[{C}_{v,m} = \frac {5}{2}R ]$ at

$300$ K and

$5$ atm is expanded adiabatically to a final pressure of

$2$ atm against a constant pressure of

$2$ atm. The final temperature (in Kelvin) of the gas is:

0%
$270.2$
0%
$273.0$
0%
$248.5$
0%
$200.5$

Explanation

Since the process is an adiabatic processs,

$q = 0$ and hence,

$\Delta U = w$

$n{C}_{v,m}({T}_{2} - {T}_{1}) = - {P}_{ext} \left [\dfrac{nR{T}_{2}}{{P}_{2}} - \dfrac{nR{T}_{1}}{{P}_{1}}\right ]$

For one mole, we have

${C}_{v,m}({T}_{2} - {T}_{1}) = {P}_{ext} R \left [\dfrac {{T}_{1}}{{P}_{1}} - \dfrac {{T}_{2}}{{P}_{2}} \right ]$

$\dfrac{5}{2} R ({T}_{2} - 300) = 2 \times R \left [\dfrac {300}{5} - \dfrac {{T}_{2}}{2} \right ]$

$\implies {T}_{2} \approx 248.5\ K$

Hence, the final temperature of the gas is

$248.5\ K$ .

Maxwell"s velocity distribution curve is given for the same quantity for two different temperatures. For the given curves then which of the folowing relation is true?

0%
$T_{1}> T_{2}$
0%
$T_{1}< T_{2}$
0%
$T_{1}\leq T_{2}$
0%
$T_{1}= T_{2}$

Explanation

Most probable velocity

$v_{mp} = \sqrt{\dfrac{2kT}{m}}$
As T increases

$v_{mp}$ increases.

$(v_{mp})_{T_2} \gt (v_{mp})_{T_1}$

$\Rightarrow T_2 \gt T_1$

$\Rightarrow T_1 \lt T_2$

During an experiment, an ideal gas is found to obey a condition

$\dfrac{P^{2}}{\rho }$

$=$ constant [

$\rho =$ density of the gas]. The gas is initially at temperature T, pressure P and density

$\rho$ . The gas expands such that density changes to

$\dfrac{\rho}{2}$

0%
The pressure of the gas changes to $\sqrt{2}P$
0%
The temperature of the gas changes to $\sqrt{2}T$
0%
The graph of the above process on the P-T diagram is parabola.
0%
The graph of the above process on the P-T diagram is hyperbola.

Explanation

Equation of process

$\Rightarrow \dfrac{P^{2}}{\rho }=$ constant

$= C$ .... (1)
Equation of state

$\Rightarrow \dfrac{P}{\rho }=\dfrac{R}{M}T$ .... (2)
From 1 and 2,

$PT=$ constant

$\Rightarrow$ C is false, D is true.
Now,

$\rho$ -changes to

$\dfrac{\rho }{2}\Rightarrow$ P changes to

$\dfrac{P}{\sqrt{2}}$ from equation (1)

$\Rightarrow$ A is false.
Hence, T changes to

$\sqrt{2}T \Rightarrow$ B is true

A monoatomic ideal gas (

${C}_{V} = \dfrac {3}{2} R$ ) is allowed to expand adiabatically and reversibly from initial volume of 8L at 300 K to a volume of

${V}_{2}$ at 250 K.

${V}_{2}$ is:

(Given

${(4.8)}^{1/2}$ = 2.2)

0%
10.5 L
0%
23 L
0%
8.5 L
0%
50.5 L

Explanation

$T{V}^{\gamma - 1} = constant$

$\gamma = \dfrac{5}{3}$

Therefore,

$\gamma - 1 = \dfrac{2}{3}$

$300 \times {(8)}^{2/3} = 250 \times {({V}_{2})}^{2/3} \implies {({V}_{2})}^{2/3} = 4.8$

$\implies {V}_{2} = {(4.8)}^{3/2} \approx 4.8 \times 2.2 = 10.5 L$

On heating 128 g of oxygen gas from 0

$^{\circ}$ C to 100

$^{\circ}$ C,

$C_V$ and

$C_P$ on an average are 5 and 7 cal mol

$^{-1}$ degree

$^{-1}$ , the value of

$\Delta$ U and

$\Delta$ H are respectively :

0%
2800 cal, 2000 cal
0%
2000 cal, 2800 cal
0%
280 cal, 200 cal
0%
none of the above

Explanation

128 g of oxygen (molecular weight = 32) corresponds to 4 moles.

The enthalpy change is

$\Delta H = n C_P \Delta T=4 \times 7\times 100=2800$ cal.

The change in internal energy is

$\Delta U = n C_V \Delta T=4\times 5\times 100=2000$ cal.

The heat capacity of a certain amount of a particular gas at constant pressure is greater than that at constant volume by 29.1 J/K. Match the items given in Column I with the items given in Column II.

List 1	List 2
If the gas is monatomic, heat capacity at constant volume	131 J/K
If the gas monatomic, heat capacity at constant pressure	43.7 J/K
If the gas is rigid diatomic, heat capacity at constant pressure	72.7 J/K
If the gas is vibrating diatomic, heat capacity at constant pressure	102 J/K

0%
A $\rightarrow$ 2, B $\rightarrow$ 3, C $\rightarrow$ 4, D $\rightarrow$ 1
0%
B $\rightarrow$ 1, C $\rightarrow$ 3, D $\rightarrow$ 4, A $\rightarrow$ 1
0%
C $\rightarrow$ 1, D $\rightarrow$ 3, A $\rightarrow$ 4, B $\rightarrow$ 1
0%
D $\rightarrow$ 1, A $\rightarrow$ 3, B $\rightarrow$ 4, C $\rightarrow$ 1

Explanation

Given:

$C_p-C_v= 29.1$ ....(1)
For monoatomic gas,

$\frac{C_p}{C_v}=\frac{5}{3}$ ...(2)
Substituting for

$C_p$ in (1), we get

$\frac{2}{3}C_v= 29.1$

$\therefore C_v= 43.7 \:J\:K^{-1}$

$C_p= 43.7 +29.1 =72.9 \:J\:K^{-1}$
If the gas is diatomic

$C_v=\frac{5}{7}C_p$

$\therefore \frac{2}{7}C_p= 29.1$

$\therefore CP=102 \:J\:K^{-1}$

If the gas is vibrating di-atomic

$C_v= \frac{7}{9}C_p$

$\therefore \frac{2}{9}C_p= 29.1$

$\therefore C_p=131 \:J\:K^{-1}$

$S_1$ : At extermely high temperature degree of freedom of monatomic gas becomedue to activation of vibrational degree of freedom.

$S_2$ : For a uniformly charged solid non-conducting sphere potential is maximum at it's centre.

$S_3$ : Breaking stress of wire depends on its corss-sectional Area.

$S_4$ : Only time varying magnetic field produces induced electric field which have lines of force.

0%
FFFT
0%
TTTF
0%
FTFT
0%
TFTF

Explanation

${S}_{1}$ : monatomic gases can never have degree of freedom more than 3.
false

${S}_{2}$ :uniformly charged solid non-conducting sphere's potential will be maximum at it's centre.(shown in image)
true

${S}_{3}$ :Breaking stress of wire doesn't depends on its corss-sectional Area but breaking load of wire depends on the area of cross section of wire.
false

${S}_{4}$ :Only time varying magnetic field can induce electric field which have lines of force.
true
answer is C.

At what temperature does the mean kinetic energy of hydrogen atoms increase to such an extent that they will escape out of the gravitational field of earth forever?

0%
$10075^o$ K
0%
$10000^o$ K
0%
$12075^o$ K
0%
$20000^o$ K

Explanation

When the velocity of the hydrogen atoms become equal to escape velocity of earth, then they will be able to escape from earth.

Escape velocity of Earth=11200m/s

RMS velocity of Hydrogen atom=

$\sqrt{\dfrac{3 \times RT} {M}}$

$R=8.314JK^{-1}mol^{-1}$

$M=0.002Kg$

Substituting we get

$T=10075K$

Consider a classroom of dimensions

$(5 \times 10 \times 3)\$ m

$^3$ at temperature

$20^{o}$ C and pressure

$1$ atm. There are

$50$ people in the room, each losing energy at the average of

$150$ watts. Assuming that the walls, ceiling, floor, and furniture are perfectly insulated and none of them is absorbing heat. How much time will be needed for raising the temperature of air in the room to the body temperature (37

$^{o}C$ )? [For air

$C_p = \dfrac{7}{2} R$ and neglect the loss of air to the outside as the temperature rises]

0%
$422$ sec
0%
$411.3$ sec
0%
$421.1$ sec
0%
$413.1$ sec

Explanation

The volume of air present in the room is

$5 \times 10 \times 3 = 150 \ m^3 = 150 \times 10^3$

$litre$ (

$\because\ 1\ m^3=1000\ L$ )

The number of moles of air is calculated using the ideal gas equation.

$n= \dfrac{PV}{RT} = \dfrac{1 \times 150 \times 10^3}{0.0821 \times 293} = 6.236 \times 10^3$

For 1 mole of air,

$\left ( \dfrac{\partial H}{\partial T} \right )_P = C_p = \dfrac{\Delta H }{\Delta T}$

For n moles of air,

$\therefore \Delta H = n \times C_p \times \Delta T$

Substitute values in the above expression,

$\Delta H= 6.236 \times 10^3 \times \dfrac 72 \times 8.314 \times (310 - 293)= 3.085 \times 10^6 J$

Thus, the heat need to increase the temperature of the room to

$37^{\circ}$ C is

$=3.085 \times 10^6 J$

The heat released by

$50$ people is

$=150 \times 50\ J/sec=7500 \ J/sec$ . This is the heat provided in 1 second.

$\therefore$ The time required to provide

$3.085 \times 10^6 J$ heat is

$\displaystyle \dfrac{1 \times 3.085 \times 10^6}{7500} = 411.3$ sec.

Hence, the correct option is

$\text{B}$

$1$ mole of an ideal gas

$A\ ({ C }_{ v,m }=3R)$ and

$2$ moles of an ideal gas

$B\ \left( { C }_{ v,m }=\cfrac { 3 }{ 2 } R \right)$ are taken in a container and expanded reversibly and adiabatically from

$1$ litre to

$4$ litres starting from initial temperature of

$320K$ .

$\Delta E$ or

$\Delta U$ for the process is:

0%
$-240R$
0%
$240R$
0%
$480R$
0%
$-960R$

In a certain gas

$\displaystyle \frac{2}{5}$ th of the energy of molecules is associated with the rotation of molecules and the rest of it is associated with the motion of the centre of mass. The average translation energy of one such molecule, when the temperature is

$27^{\circ}C$ is given by

$x\times 10^{-23}\ J$ ,then find

$x$ ?

0%
621
0%
623
0%
6.21
0%
62.1

Explanation

Since two fifth energy is associated with the rotation, thus there are five degrees of freedom out of which two are rotational and three translational.

Hence,

${ E }_{ trans. }=\dfrac { 3 }{ 2 } kT=\dfrac { 3 }{ 2 } (1.38\times { 10 }^{ -23 })(300)=6.21\times { 10 }^{ -21 }J$

Answer is

$621\times { 10 }^{ -23 }J$

One box containing

$1$ mole of

$He$ at

$7/3 T_0$ and other box containing

$1$ mole of a polyatomic gas

$(\gamma=1.33)$ at

$T_0$ are placed together to attain thermal equilibrium. The final temperature becomes

$T_f$ . Then :

0%
$T_f=\dfrac {9}{13}T_0$
0%
$T_f=\dfrac {13}{9}T_0$
0%
$T_f=\dfrac {5}{2}T_0$
0%
$T_f=\dfrac {3}{2}T_0$

Explanation

At thermal equilibrium, both have the same temperature and heat given by one gas equal to the heat absorbed by other gas.

Heat released by

$He=$ Heat taken up by polyatomic gas

$1\times C_v\times \Delta T=1\times C_v\times \Delta T$

$1\times \dfrac {3R}{2}(7/3T_0-T_f)=1\times 3R\times (T_f-T_0)$

$\left [C_v=\dfrac {3}{2}\ R\ \text {for He and}\ C_v=\frac {3}{2}R+X=\underset {\text {for polyatomic gas}}{\dfrac {3}{2}R+\dfrac {3}{2}R=3R}\right ]$

$\therefore \dfrac {7}{2}T_0-\dfrac {3}{2}T_f=3T_f-3T_0$

$\dfrac {13}{2}T_0=\dfrac {9}{2}T_f$

$\therefore T_f=\dfrac {13}{9}T_0$

Identify the type of gas filled in container A and B respectively

0%
Mono, mono
0%
Dia, dia
0%
Mono, dia
0%
Dia, Mono

Explanation

Since the temperature rise and heat supplied to both the container is same.

$Q=n_AC_{VA}\Delta T=n_BC_{VB}\Delta T$

$\Rightarrow n_AC_{VA}=n_BC_{VB}\ -(1)$

also,

$\Delta P_{A}V=n_A R\Delta T\ -(2)$

and,

$\Delta P_{B}V=n_B R\Delta T\ -(3)$

Dividing

$(2)$ by

$(3)$ ,

$\dfrac{n_A}{n_B}=\dfrac{\Delta P_A}{\Delta P_B}=2.5/1.5=5/3$

$\Rightarrow n_A=5n_B/3$

From equation

$(1)$

$C_{VB}/C_{VA}=5/3=(5R/2)/(3R/2)$

$C_{VB}=5R/2$ , so

$B$ is diatomic.

$C_{VA}=3R/2$ , so

$A$ is Mono-atomic.

In a certain gas

$\displaystyle \frac{2}{5}$ th of the energy of molecules is associated with the rotation of molecules and the rest of it is associated with the motion of the centre of mass. How much energy must be supplied to one mole of this gas at constant volume to raise the temperature by

$1^{\circ}C$ ?

0%
$20.8 J$
0%
$28.8 J$
0%
$26.8 J$
0%
$25.8 J$

Explanation

Since two fifth energy is associated with the rotation, thus there are five degrees of freedom out of which two are rotational and three translational.

Thus

${ e }_{ total }=\dfrac { 5 }{ 2 } RT$

$\Delta E=\dfrac { 5 }{ 2 } (8.314)1=20.8J$

A rigid container of negligible heat capacity contains one mole of a gas. 10 calories of heat is required to increase the temperature by

$2^oC$ in the Boyle's temperature range. The gas in container may be

0%
He
0%
$H_2$
0%
$O_2$
0%
Ar

Explanation

As we know,

$\Delta H =nC_v \Delta T$

$10 = 1\times C_v \times 2$

Therefore,

$C_v= 5$

$C_p - C_v = R$ , and

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

Also,

$C_v= \dfrac{R}{\gamma -1} = 5 \implies \gamma -1 = \dfrac{2}{5}$

$\gamma =1.4$

So, gas is diatomic.

A monoatomic ideal gas undergoes a process in which the ratio of P to V at any instant is constant and equal to unity. The molar heat capacity of gas is:

0%
$\frac {4R}{2}$
0%
$\frac {3R}{2}$
0%
$\frac {5R}{2}$
0%
zero

Explanation

$\textbf{Given:-}$ Mono-atomic ideal gas

$\dfrac{p}{V} = 1$

$\textbf{Solution:-}$

The molar heat capacity for any process is given by a following expression

$C = C_{V} + \dfrac{R}{1 - \gamma } when pV^{\gamma } = constant$

$\dfrac{C_{P}}{C_{V}} = \gamma$

$\dfrac{p}{V} = 1 ie, pV^{-1} = constant$

$C = \dfrac{3}{2}R + \dfrac{R}{1 - (-1)} = \dfrac{3}{2}R + \dfrac{R}{2} = \dfrac{4}{2}R$

$\textbf{Hence the correct option is A}$

Which of the following statements is wrong about the gas?

0%
The gas is monoatomic
0%
The gas is $He$
0%
The gas is a rare gas or inert gas
0%
The number of molecules of gas is $3.01 \times 10^{24}$

Explanation

Given,

$C_p=0.125 \cal/g, C_v = 0.075\ cal/g$

Also as we know,

$C_p-C_v = \dfrac {R}{M}$

$M=\dfrac {2}{0.125-0.075}=40$ so molar mass of gas is 40.

Also,

$\dfrac {C_p}{C_v}=\dfrac {0.125}{0.075}=\dfrac {5}{3}=1.66$ .

Therefore, gas is monoatomic and also the molar mass of gas is 40

$g/mol$ .

Moles of gas

$X$ =

$5$ moles

$= 5 \times 6.023 \times 10^{23} = 3.01 \times 10^{24}$ molecules.

Thus, gas is

$Ar$ which is an inert gas.

Select the correct orders for gases represented with their characteristics:

0%
Critical temperature < Boyle's temperature
0%
van der Waal's constant 'a' : $H_2O > NH_3 > N_2 > Ne$
0%
Heat capacity: $NH_3 > CO_2 > O_2 = N_2 > Ar$
0%
Mean free path: $He > H_2 > O_2 > N_2 > CO_2$

Explanation

As we know,

$C_v= R/\gamma -1$
so as atomcity of gas increases, heat capacity increases.
so order is
Heat capacity:

$NH_3 > CO_2 > O_2 = N_2 > Ar$
Also as atomcity and mass increases, mean path decreases so order is
Mean free path:

$He > H_2 > O_2 > N_2 > CO_2$
The 'a' values for a given gas are measure of intermolecular forces of attraction. More are the intermolecular forces of attraction, more will be the value of a.
so order is
van der Waal's constant 'a' :

$H_2O > NH_3 > N_2 > Ne$

Figure shows the variation of the internal energy

$U$ with density

$\rho$ of one mole of an ideal monatomic gas for thermodynamic cycle ABCA. Here process AB is a part of rectangular hyperbola

0%
process AB is isothermal & net work in cycle is done on gas
0%
process AB is isothermal & net work in cycle is done by the gas
0%
process AB is isobaric & net work in cycle is done on the gas
0%
process AB is adiabatic & net work in cycle is done by gas

Explanation

The internal energy is given by

$U = C_V T$

A rectangular hyperbola in

$x$ and

$y$ axes is given by

$xy = \textrm{constant}$ .

Thus, the leg

$AB$ in the cycle corresponds to

$\rho U = \textrm{constant}$ , i.e.,

$\rho C_VT = \textrm{const}$ ....(1)

We also know

$P =\rho RT$ .....(2)

From (1) and (2), we get

$P = \textrm{constant}$

Thus, the process AB is "Isobaric".

In the graph, from A to B, the internal energy decreases. Thus, the temperature decreases.

Hence, the density increases, i.e., volume decreases.

Work done by the gas is

$W = P\Delta V$

As,

$\Delta V$ is negative, work done by the gas is negative. i.e., Work is done on the gas.

Hence correct answer is option C.

Keeping the number of moles, volume and pressure the same, which of the following are the same for all ideal gas?

0%
rms speed of a molecule
0%
density
0%
temperature
0%
average of magnitude of momentum

Explanation

Ideal gas equation

$\rightarrow PV=nRT$

Given that

$n_1V_1P$ are same.

So,

$T$ i.e, temperature is same for all ideal gas.

Hence, the answer is Temperature.

A sample of a fluorocarbon was allowed to expand reversibly and adiabatically to twice its volume. In the expansion the temperature dropped from 298.15 to 248.44 K. Assume the gas behaves perfectly. Estimate the value of

$C_{V_1 m}$ .

0%
$C_{V_1m} = 31.6 Jk^{1} mol^{1}$
0%
$C_{V_1m} = 3.16 Jk^{1} mol^{1}$
0%
$C_{V_1m} = 316 Jk^{1} mol^{1}$
0%
None of these

Explanation

Process reversibly adiabatic

$T_1 = 298.15 K {\;} V_2 = 2V_1$

$T_2 = 248.44 K {\;} Pv^{\gamma}= K {\;} PV = nRT$

$\dfrac {T}{V}\cdot V^{\gamma} K T_1V_1^{\gamma1} = T_2V_2^{\gamma1}$

$\left (\dfrac {T_1}{T_2}\right )=\left (\dfrac {V_2}{V_1}\right )^{\gamma-1} {\;} \left (\dfrac {298.15}{248.44}\right )=2^{\gamma-1}$

$1.2=2^{\gamma-1} {;} log 1.2=log 2. (\gamma-1)$

$\gamma-1=\dfrac {log 1.2}{10g 2} {\;} \gamma-1=0.263$

Now

$nC_v(T_2-T_1)=\dfrac {P_2V_2-P_1V_1}{\gamma-1}$

$C_{V_1m}=\left (\dfrac {R}{\gamma-1}\right )=\dfrac {nR(T_2-T_1)}{(\gamma-1)}$

$C_{V_1m}=\dfrac {8.314}{0.263} {\;} C_{V_1m}=31.61Jk^{1} mol^{1}$

One mole of an ideal gas

$\left (C_{v,m}\, =\, \displaystyle \frac {5}{2} R \right )$ at

$300$ K and

$5$ atm is expanded adiabatically to a final pressure of

$2$ atm against a constant pressure of

$2$ atm. Final temperature of the gas is:

0%
$270$ K
0%
$273$ K
0%
$248.5$ K
0%
$200$ K

Explanation

For adiabatic expansion against a constant external pressure:

$W = ‐P_{ex}\Delta V = C_v\Delta T$ ,

$P_{ex}(V_f – V_i) = C_v(T_f – T_i)$

$C_v T_i + P_{ex} V_i = C_v T_f + P_{ex} Vf$

Also,

$V_f= d\dfrac {nRT_f}{P_{ex}} , \quad \quad \quad V_i= \dfrac {nRT_i}{P_i}$

$C_v T_i + nR T_i \left ( \dfrac{P_{ex}}{P_i } \right) = C_v T_f + P_{ex} \left (\dfrac{nR T_f}{P_{ex}} \right ) = T_f [C_v + nR]$

$T_f = T_i \times \dfrac{ \left [ \dfrac{5}{2}n R + nR \left ( \dfrac{P_{ex} }{P_i } \right ) \right ]}{[\frac{5}{2}nR + nR]} = T_i \times \dfrac {\left [\frac{5}{2} + \left ( \dfrac{P_{ex}}{P_i } \right ) \right ]}{\left (\dfrac{5}{2} + 1 \right )}$

$T_f= 300 \times \dfrac {\left [ \dfrac{5}{2}+ \dfrac{2}{5} \right ]} {\left [\dfrac{7}{2} \right ]} = 300\times 0.828 = 248.5K$

Hence, the correct option is

$C$

The quantity

$\dfrac {2U}{fkT}$ represents (where

$U=$ internal energy of gas)

0%
mass of the gas
0%
kinetic energy of the gas
0%
number of moles of the gas
0%
number of molecules in the gas

Explanation

According to the law of Equipartition of energy to the molecule of a gas states that for each degree of freedom(f) for a molecule the associated energy is

$\dfrac{1}{2}KT$

So, let

$U$ be the total energy of all the molecules

$n$ each having degree of freedom equal to

$1.$

So,

$U=nf\dfrac{1}{2}KT$

$\Rightarrow \dfrac{2U}{fKT}=n$

Hence, the answer is number of molecules in the gas.

In the figure, an ideal gas is expanded from volume

$V_0$ to

$2V_0$ under three different processes. Process 1 is isobaric, process 2 is isothermal and process 3 is adiabatic. Let

$\Delta U_1, \Delta U_2$ and

$\Delta U_3$ be the change in the internal energy of the gas is these three processes, Then:

0%
$\Delta U_1 > \Delta U_2 > \Delta U_3$
0%
$\Delta U_1 < \Delta U_2 < \Delta U_3$
0%
$\Delta U_2 < \Delta U_1 < \Delta U_3$
0%
$\Delta U_2 < \Delta U_3 < \Delta U_1$

Explanation

According to first law of thermodynamic

$dQ=dU+P\triangle V,$

Isobaric Isothermal Adiabatic

Pressure is constant.

$\triangle T=0$

$dQ=0$

So,

$dQ-P\triangle V=\triangle V_1$

$\triangle U_2=0$

$-P\triangle V=\triangle U_3$

So,

$\triangle U_1>\triangle U_2>\triangle U_3$

Hence, the answer is

$\triangle U_1>\triangle U_2>\triangle U_3.$

Among the following, identify the substance in which molecules possess vibratory, rotatory and translatory motion, but their movements are not random.

0%
Bromine
0%
Iodine
0%
Ammonia
0%
Silicon dioxide

Name

$A, B, C, D,E,$ and

$F$ in the following diagram showing change of state

0%
A $=$ Melting, B $=$ Vaporisation, C $=$ Condensation, D $=$ Solidification, E $=$ Sublimation, F $=$ Solidification of gaseous state
0%
A $=$ Sublimation, B $=$ Vaporisation, C $=$ Condensation, D $=$ Solidification, E $=$ Melting, F $=$ Solidification of gaseous state
0%
A $=$ Condensation, B $=$ Vaporisation, C $=$ Melting, D $=$ Solidification, E $=$ Sublimation, F $=$ Solidification of gaseous state
0%
A $=$ Vaporisation, B $=$ Melting, C $=$ Condensation, D $=$ Solidification, E $=$ Sublimation, F $=$ Solidification of gaseous state

If one mole of a monoatomic gas

$\left (\gamma = \dfrac {5}{3}\right )$ is mixed with one mole of diatomic gas

$\left (\gamma = \dfrac {7}{5}\right )$ , the value of

$\gamma$ for the mixture is :

0%
$1.40$
0%
$1.50$
0%
$1.53$
0%
$3.07$

Explanation

For monoatomic gas :

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

$\displaystyle C_p = \dfrac {5}{2}R$

For diatomic gas :

$\displaystyle C_v = \dfrac {5}{2}R$

$\displaystyle C_p = \dfrac {7}{2}R$

For mixture of 1 mole of each :

$\displaystyle C_v = \dfrac {\dfrac {3}{2}R+\dfrac {5}{2}R}{2}=2R$

$\displaystyle C_p = \dfrac {\dfrac {5}{2}R+\dfrac {7}{2}R}{2}=3R$

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

$\displaystyle \gamma = \dfrac {3R}{2R}$

$\displaystyle \gamma = 1.50$

Hence, the correct option is B.

$\displaystyle N_{2}$ gas is assumed to behave ideally A given volume of

$\displaystyle N_{2}$ originally at 373 k and 0.1013 M pa pressure is adiabatically compressed due to which its temperature rises to

$\displaystyle 673 \ K \left ( Cv=\frac{5}{2}R \right )$

Which of the following statement(s) is/are correct?

0%
The change in internal energy is 6235.5 J $\displaystyle mole^{-1}$
0%
In this case the final internal pressure is equal to the external pressure
0%
The final pressure of $\displaystyle N_{2}$ is approximately 0.38 MPa
0%
The final pressure of $\displaystyle N_{2}$ is approximately 0.02 Mpa

Explanation

$\displaystyle \Delta E=\int P_{ext}dV=-P_{ext}(V_{2}-V_{1})$

$\displaystyle (T_{2}-T_{1})=-P_{ext}\left ( \frac{RT_{2}}{P_{2}}-\frac{RT_{1}}{P_{1}} \right )$ Here,

$\displaystyle P_{ext}=P_{2}$

$\displaystyle (T_{2}-T_{1})=-P_{2}\left [ \frac{RT_{2}}{P_{2}} -\frac{RT_{1}}{P_{1}}\right ]$

$\Rightarrow \cfrac{5}{2}R(673-373)=-P_{2}\left [ \cfrac{R(673)}{P_{2}}-\cfrac{R(373)}{0.1013} \right ]$

$\displaystyle \Rightarrow \cfrac{5}{2}\times 300=-673+\frac{373P_{2}}{0.1013}$

$\therefore P_{2}=\cfrac{(750+673)}{373}\times 0.1013=0.3865M Pa$

$\displaystyle \Delta E=Cv(T_{2}-T_{1})=\frac{5}{2}R\times 300=\frac{5}{2}R\times 300=\frac{5}{2}\times 8.314\times 300J/mol=6235.5J/mol$

Hence, A, B and C are correct.

According to the Kinetic Molecular Theory, temperature is directly proportional to:

0%
average kinetic energy
0%
lattice energy
0%
ionization energy
0%
free energy
0%
average potential energy

A molecule of gas in a container hits one wall (1) normally and rebounds back. It suffers no collision and hits the opposite wall (2) which is at an angle of

$30^\circ$ with wall 1.
Assuming the collisions to be elastic and the small collision time to be the same for both the walls, the magnitude of average force by wall(

$F_2$ ) provided to the molecule during collision satisfy :

0%
$F_1\, >\, F_2$
0%
$F_1\, <\, F_2$
0%
$F_1\, =\,F_2$ , bothnon-zero
0%
$F_1\, =\, F_2\, =\, 0$

Explanation

Work done by the wall during a collision is equal to the change in momentum of the collided particle per unit time of collision.

Let the collision time in both cases be

$t$ , the mass of the particle be

$m$ and the speed be

$u$ .

Change in momentum for the collision at Wall1 is

$\Delta p_1 = mu-(-mu) = 2mu$

The final momentum of the gas molecule after its collision with wall 2 makes

$120^o$ angle with the momentum just before its collision with wall 2.

Change in momentum for the collision at Wall2 is

$\Delta p_2 = \sqrt{(mu)^2+(mu)^2-2(mu)(mu)\cos(120^\circ)} = \sqrt{3}mu$

Force is given by

$F = \dfrac{p}{t}$

Thus,

$F_1 = 2mu/t$ and

$F_2 = \sqrt{3}mu/t$ .

Hence,

$F_1>F_2$

A gaseous mixture enclosed in a vessel consists of one g mole of a gas A with

$\displaystyle \gamma =\left ( \frac{5}{3} \right )$ and some amount of gas B with

$\displaystyle \gamma = \frac{7}{5}$ at a temperature The gasses A and B do not react with each other and are assumed to be ideal Find the number of g moles of the gas B if

$\displaystyle \gamma$ for the gaseous mixture is

$\displaystyle \left ( \frac{19}{13} \right )$

0%
2
0%
5
0%
4
0%
3

Explanation

Change in internal energy of the first gas + change in internal energy of the second gas = change in internal energy of the mixture

$\mu_1 (C_{v})_{1}\Delta T+\mu _{2}(C_{v})_{2} \Delta T=(\mu _{1}+\mu _{2})(C_{v})_{mix}\Delta T$

$\mu _{1}\dfrac{R}{\gamma _{1}-1}+\mu _{2}\dfrac{R}{\gamma _{2}-1}=(\mu _{1}+\mu _{2})\dfrac{R}{\gamma _{mix}-1}; \, \therefore\, \gamma_{mix}=\dfrac{\mu _{1}\gamma _{1}(\gamma _{2}-1)+\mu _{2}\gamma _{2}(\gamma _{1}-1)}{\mu _{1}(\gamma _{2}-1)+\mu _{2}(\gamma _{1}-1)}$

$\dfrac{19}{13}=\dfrac{1\times (5/3)(7/5-1)+\mu _{2}\times (7/5)(5/3-1)}{1(7/5-1)+\mu _{2}(5/3-1)}$ or

$\mu _{2}=2$

One mole of an ideal gas in initial state A undergoes a cyclic process ABCA, as shown in the figure. Its pressure at A is

$P_0$ . Choose the correct option(s) from the following.

0%
Internal energies at A and B are the same
0%
Work done by the gas in process AB is $P_oV_o$ ln $4$
0%
Pressure at C is $\dfrac{P_o}{4}$
0%
Temperature at C is $\dfrac{T_o}{4}$

Explanation

From the graph given in the question, we can see that temperature at states Aand C are same,

i.e,

$T_A=T_B$

Internal energy

$=\dfrac{t}{2}nRT=\dfrac{t}{2}RT$ ( Here

$n=1$ )

$\Rightarrow$ Here,

$f=$ degree of freedom

$n=$ moles

$R=$ universal gas constant.

As,

$f$ and

$R$ are constant and

$T_A=T_B,$

So, internal energy of the gas at A and B are same .

Hence, option A is correct.

For the process AB, temperature is constant throughout i.e, the process is isothermal.

Work done by a gas in an isothermal process

$=nRT\iota \eta \left(\dfrac{v_2}{v_1}\right)$

$=\left(1\right) \left(R\right) \left(T_0\right) \iota \eta \left(\dfrac{v_B}{v_A}\right)$ [ Hence

$n=1$ ]

$=RT_0\iota \eta \left(\dfrac{4v_0}{v_0}\right)$ [ On applying,

$Pv=nRT$ at

$'A',$ we will get

$RT_0=p_0v_0$ ]

$=P_0v_0 \iota \eta 4.$

So, option B is correct.

Hence, both option A and B are correct.

If the radii of two copper spheres are in the ratio

$1 : 3$ and increase in their temperatures are in the ratio

$9 : 1$ then the ratio of the increase in their internal energy will be

0%
$1 : 4$
0%
$2 : 1$
0%
$4 : 1$
0%
$1 : 3$

Relation between pressure (

$P$ ) and energy density (

$E$ ) of an ideal gas is-

0%
$P=2/3E$
0%
$P=3/2E$
0%
$P=3/5E$
0%
$P=E$

Explanation

Kinetic energy

$=\dfrac{1}{2}{ MV }_{ rms }$

$\Rightarrow \dfrac{1}{2}M\left( \dfrac { 3RT }{ M } \right)$

$[M=$ molar mass,

${ V }_{ rms }=\sqrt { \dfrac { 3KT }{ { m } } } =\sqrt { \dfrac { 3RT }{ M } } ]$

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

$\Rightarrow K.E=\dfrac{3}{2}PV$

$[PV=RT]$

$\Rightarrow \dfrac{K.E}{V}=\dfrac{3}{2}P$

$\Rightarrow E=\dfrac{3P}{2}$

$E=$ Energy density.

Hence, the answer is

$P=\dfrac{2}{3}E.$

An insulated container is divided into two equal portions. One portion contains an ideal monoatomic gas at pressure

$P$ and temperature

$T$ , while the other portion is a perfect vacuum. If a hole is opened between the two portions, the change in internal energy of the gas is

0%
Zero
0%
Equal to work done by the gas
0%
Equal to work done on the gas
0%
$3RT/2$

Let

$\overline {v}, v_{rms}$ and

$v_{p}$ , respectively, denote the mean speed, root mean square speed and most probable speed of the molecules in an ideal monatiomc gas at absolute temperature

$T$ . The mass of a molecules is

$m$ . Then

0%
No molecule can have a speed greater than $\sqrt {2} v_{rms}$
0%
No molecule can have speed less than $v_p/\sqrt {2}$
0%
$v_{p} < \overline {v} < v_{rms}$
0%
The average kinetic energy of a molecule is $\dfrac {3}{4}mv^{2}_{P}$

Explanation

$v_{rms} = \sqrt {\dfrac {3RT}{M}}, \overline {v} = \sqrt {\dfrac {8}{\pi} \cdot \dfrac {RT}{M}} \approx \sqrt {\dfrac {2.5 RT}{M}}$
and

$v_{p} = \sqrt {\dfrac {2RT}{M}}$
From these expressions we can see that

$v_{p} < \overline {v} < v_{rms}$
Second,

$v_{rms} = \sqrt {\dfrac {3}{2}} v_{p}$
and average kinetic energy of a gas molecule

$= \dfrac {1}{2} mv^{2}_{rms}$

$= \dfrac {1}{2} m\left (\sqrt {\dfrac {3}{2}}v_{p} \right )^{2} = \dfrac {3}{4} mv_{p}^{2}$ .

An ideal gas is taken from the state A(pressure

$P_0$ , volume

$V_0$ ) to the state B (pressure

$P_0/2$ , volume

$2V_0$ ) along a straight line path in the P-V diagram. Select the correct statement(s) from the following.

0%
The work done by the gas in the process A to B exceeds the work that would be done by it if the system were taken from A to B along the isotherm
0%
In the T-V diagram, the path AB becomes a part of the parabola
0%
In the P-T diagram, the path AB becomes a part of hyperbola
0%
In going from A to B, the temperature T of the gas first increases to maximum value and then decreases

Explanation

Work done by the gas

$=$ area under the graph on volume axis.

Clearly, from the graph, we can see that area under the isotherm is less than the area under straight line AB process.

So, option (A) is correct.

Equation of AB :

$P=\left(\dfrac{-P_0}{2V_0}\right)V+\dfrac{3P_0}{2}$

$\Rightarrow \dfrac{nRT}{V}-\left(\dfrac{-P_0}{2V_0}\right)V+\dfrac{3P_0}{2}$ [ Applying

$PV=nRT$ ]

$\Rightarrow T=\left(\dfrac{-P_0}{2nRV_0}\right)V^2+\dfrac{3P_0}{2}V \equiv y=-Ax^2+Bx$

On drawing various isotherms across AB, we can see that temperature first increases then decreases.

Hence, this question has multiple answers.

Volume versus temperature graphs for a given mass of an ideal gas are shown in figure at two different values of constant pressure. What can be inferred about relation between

$P_1$ and

$P_2$ ?

0%
$P_1 > P_2$
0%
$P_1 = P_2$
0%
$P_1 < P_2$
0%
Data is insufficient

Explanation

As pressure and quantity of gas in system are contant. So, by ideal gas equation

$PV=nRT$

$V\propto T$ as

$n,R$ and

$P$ are constant

$\cfrac{{V}_{1}}{{T}_{1}}=$ costant or slope of graph is constant

$V=\cfrac{nRT}{P}$

$\cfrac{dV}{dT}=\cfrac{nR}{P}$ so

$\cfrac{dv}{dT}$ increase when

$P$ decreases

$\cfrac{dV}{dT}\propto \cfrac{1}{P}$
as slope of

${P}_{1}$ is smaller than

${P}_{2}$ . Hence

${P}_{1}> {P}_{2}$ verifies option (a)

The equation of state for 5 g of oxygen at a pressure P and temperature T, when occupying a volume V, will be

0%
$PV=\Big(\dfrac{5}{32}\Big)RT$
0%
$PV = 5 RT$
0%
$PV = \Big(\dfrac{5}{2}\Big)RT$
0%
$PV = \Big(\dfrac{5}{16}\Big)RT$

Explanation

Given,

$m=5g$

$M=32g$ (molecular mass of the oxygen molecule)

Number of moles,

$n=\dfrac{m}{M}$

$n=\dfrac{5}{32}$

From the equation of gas,

$PV=nRT$

$PV=(\dfrac{5}{32})RT$

The maximum high temperature molar heat capacity at constant volume to be expected for acetylene which is linear molecule is:

0%
9 cal/deg-mol
0%
12 cal/deg-mol
0%
19 cal/deg-mol
0%
14 cal/deg-mol

In the given figure conainer

$A$ holds an ideal gas at a pressure of

$5.0 \times 10^{5}\ Pa$ and a temperature of

$300\ K.$ It is connected by the tin tube (and a closed valve) to container

$B$ , with four times the volume of

$A$ . Container

$B$ hold same ideal gas at a pressure of

$1.0\ \times 10^{5}\ Pa$ and a temperature of

$400\ K$ . The valve is opened to allow the pressure to equalize, but the temperature of each container kept constant at its initial value. The final pressure in the two containers will be close to :

0%
$2.0 \times 10^{5}\ Pa$
0%
$2.5 \times 10^{5}\ Pa$
0%
$3.0 \times 10^{5}\ Pa$
0%
$3.5 \times 10^{5}\ Pa$

Liquid is filled in a vessel which is kept in a room with temperature

$20^\circ C$ . When the temperature of the liquid is

$80^\circ C$ , then it losses heat at the rate of 60 cal/sec.What will be the rate of loss of heat when the temperature of the liquid is

$40^\circ C$ .

0%
180 cal/sec
0%
40 cal/sec
0%
30 cal/sec
0%
20 cal/sec

Explanation

We know rate of heat transfer

$q= h A\Delta T$

$q= C\Delta T$

where C is constant

$60= C.(80-20)$

$60= C\times 60$

$c=1$

for

$q= 40^\circ$

$q=c (40-20)$

$=1\times 20$

$q=20 cal/sec$

CBSE Questions for Class 11 Engineering Physics Kinetic Theory Quiz 14 - 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 14 - MCQExams.com

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Practice Class 11 Engineering Physics Quiz Questions and Answers

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