MCQ Questions for Class 11 Engineering Physics Kinetic Theory Quiz 7

'N' molecules each of mass 'm', of gas A and '2N' molecules, each of mass '2m', of gas B are contained in the same vessel. Which is maintained at a temperature T. The mean square of the velocity of molecules of B type is denoted by

$V^{2}$ and the mean square of the X component of the velocity of A type is denoted by

$\omega^{2}$ ,

$\displaystyle\ \frac{\omega^{2}}{V^{2}}=$

0%
2
0%
1
0%
1/3
0%
2/3

Explanation

Total KE of A type of molecules =

$\displaystyle\ \frac{3}{2} m\omega^{2}$
Total KE of A type of molecule is

$K.E_{A}$ =

$\displaystyle\ \frac{1}{2}[(V_{r.ms})^{2}_{x}+(V_{rms})^{2}_{y}+(V_{rms})^{2}_{z}]$
but

$(V_{rms})_{x}$ =

$\omega$
So

$(V_{rms})_{y}$ =

$(V_{rms})_{z}$ =

$\omega$
Total KE of B type molecules

$= \displaystyle\ \frac{1}{2}\times2mv^{2}$ =

$m.v^{2}$
Now,

$\displaystyle\ \frac{3}{2}\times m\omega^{2}$ =

$m v^{2}$
or

$(\omega^{2}/v^{2})$ =

$\displaystyle\ \frac{2}{3}$

A vessel has 6g of hydrogen at pressure P and temperature 500K. A small hole is made in it so that hydrogen leaks out. How much hydrogen leaks out if the final pressure is

$P/2$ and temperature falls to 300K?

0%
2g
0%
3g
0%
4g
0%
1g

Explanation

$PV$ =

$\displaystyle\ \frac{m}{M}RT$
Initally,

$PV$ =

$\displaystyle\ \frac{6}{M}R\times500$
Finally,

$\displaystyle\ \frac{P}{2}V$ =

$\displaystyle\ \frac{(6-x)}{M}R\times300$ ( if x g gas leaks out)
Hence,

$2 = \displaystyle\ \frac{6}{6-x}\times\displaystyle\ \frac{5}{3}$

$\therefore x = 1$ gram

Graph of specific heat at constant volume for a monatomic gas is:

Explanation

For an ideal gas specific heat

$C_v$ is independent of temperature and for diatomic molecules

$C_v = \dfrac{3}{2}R$ for all temperature conditions.

So the graph between

$C_v$ and T will be a straight line parallel to the x-axis.

A graph is plotted with PV/T on y-axis and mass of the gas along x-axis for different gases. The graph is

0%
a straight line parallel to x-axis for all the gases
0%
a straight line passing through origin with a slope having a constant value for all the gases
0%
a straight line passing through origin with a slope having different values for different gases
0%
a straight line parallel to y-axis for all the gases

Explanation

$\displaystyle\ \frac{PV}{T}$ = nR =

$\left( \displaystyle\ \frac{m}{M}\right)R$
Or,

$\displaystyle\ \frac{PV}{T} = \left (\displaystyle\ \frac{R}{M}\right)m$
i.e.,

$\displaystyle\ \frac{PV}{T}$ versus on graph is straight line passing through origin with slope R/M. i.e., the slope depends on molecular mass of the gas M and is different for different gases

A gas mixture consists of 2 moles of oxygen and 4 moles of Argon at temperature T. Neglecting all vibrational moles, the total internal energy of the system is

0%
4RT
0%
15RT
0%
9RT
0%
11RT

Explanation

Internal energy of 2 moles of oxygen

$U_{2}$ =

$\mu\left(\displaystyle\ \frac{5}{2} RT\right)$

$=2. \displaystyle\ \frac{5}{2}RT$ =

$5RT$
Internal energy of 4 moles of Argon.

$U_{AR} = \mu \left(\displaystyle\ \frac{3}{2}RT\right)$

$=4. \displaystyle\ \frac{3}{2}RT$ =

$6RT$

$therefore$ Total internal energy

$U = U_{O_{2}}+ U_{AR}$

$=11RT$

If pressure of a gas contained in a closed vessel is increased by 0.4% when heated by

$1^{0}C$ , the initial temperature must be

0%
250K
0%
$250^{0}C$
0%
2500K
0%
$25^{0}C$

Explanation

We know that

$\displaystyle\ \frac{P_{1}}{T_{1}}$ = constant for constant volume.

Let the initial pressure be

$P_1$ and initial temperature be

$T_1$ .

Thus final pressure

$P_2 =1.004 P_1$ and final temperature

$T_2 = T_1 + 1$ .

Using

$\dfrac{P_1}{T_1} = \dfrac{P_2}{T_2}$
So,

$\displaystyle\ \frac{P_{1}}{T_{1}}$ =

$\displaystyle\ \frac{1.004 P_{1}}{T_{1}+1}$

$T_1 + 1 = 1.004 T_1$

$0.004 T_1 = 1$

$\Rightarrow T_1 =250K$

An ideal gas is found to obey an additional law

$VP^{2}$ = constant. The gas is initially at temperature T and volume V. When it expands to a volume 2V, the temperature becomes :

0%
$\displaystyle\ \frac{T}{\sqrt{2}}$
0%
$2T$
0%
$\displaystyle\ \frac{2T}{\sqrt{2}}$
0%
$4T$

Explanation

We are given an ideal gas which is following an additional law,

$PV^2=$ a constant ....................(1)

As the gas is ideal, the gas law is

$PV=nRT$

Taking the value of

$P$ from the above equation, we get

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

Putting this value in equation (1), we get

$\dfrac{nRT(V)^2}{V} =$ constant

$n$ and

$R$ are constant, above equation becomes,

$TV=$ a constant .....................(2)

Now, the gas is expanding from initial volume

$V_1$ to final volume

$V_2$ by

$2$ units, it becomes

$V_2=2V_1$

From equation (2), we get

$T_1 V_1 = T_2 V_2$

$T_1 V_1 = T_2 (2V_1)$

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

So the correct answer is

$\dfrac{T}{2}$

Final volume of the system will be nearly -

0%
$6.2m^{3}$
0%
$2.8m^{3}$
0%
$4.5m^{3}$
0%
$1.4m^{3}$

Explanation

$PV = nRT$
Initial volume

$= \displaystyle\ \frac{ 1.45\times10^{3}}{29} \times \displaystyle\ \frac{8.3\times300}{1.5\times10^{5}}m^{3}$

$\displaystyle\ \frac{V_{i}}{P_{i^{2}}}$ =

$\displaystyle\ \frac{V_{f}}{P_{f}^{2}}$

$\Rightarrow$

$V_{f}$ =

$\left (\displaystyle\ \frac{3.5}{1.5}\right)^{2} V_{i}$ =

$4.5 m^{3}$

$T = \displaystyle\ \frac{P_{f}V_{f}}{nR} = 3800K$

$\triangle U = nC_{V}\triangle T = n \displaystyle\ \frac{5R}{2} \triangle T = 3.6\times10^{6}J$

The heat capacity at constant volume of a sample of a monoatomic gas is

$35\ J/K$ . Find the number of moles.

0%
$12.81 \ \ mol$
0%
$21.81 \ \ mol$
0%
$4.81 \ \ mol$
0%
$2.81 \ \ mol$

Explanation

For monoatomic gas, degrees of freedom is 3.

Since

${ C }_{ V }=\dfrac { f }{ 2 } nR$

Hence,

$35=\dfrac { 3 }{ 2 } n(8.314)$

$n=\dfrac { 70 }{ 3\times 8.314 } =2.81mol$

Answer is

$2.81mol.$

If the kinetic energy of the molecules in

$5$ litres of helium at

$2$

$atm$ is

$E$ . What is the kinetic energy of molecules in

$15$ litres of oxygen at

$3$

$atm$ in terms of

$E$ ?

0%
$7.5E$
0%
$7E$
0%
$8.5E$
0%
$8E$

Explanation

Kinetic energy of n moles is given by,

$K=\dfrac { f }{ 2 } nRT=\dfrac { f }{ 2 } PV$

Thus,

$E=\dfrac { 3 }{ 2 } (5)(2)=15$

$E'=\dfrac { 5 }{ 2 } (15)(3)=112.5=7.5E$

Answer is

$7.5E$

The average kinetic energy of the molecules of an ideal gas at

$10^{\circ}C$ has the value E. The temperature at which the kinetic energy of the same gas becomes 2E is

0%
$5^{\circ}C$
0%
$10^{\circ}C$
0%
$40^{\circ}C$
0%
None of these

Explanation

The average kinetic energy of gas molecules

$T$

Thus

$\dfrac{E_2}{E_1}=\dfrac{T_2}{T_1}=2$

$\implies T_2=2\times T_1$

$=2\times 283K=566K=293^{\circ}C$

Choose the wrong options

0%
Translation kinetic energy of all ideal gases at same temperature is same
0%
In one degree of freedom all ideal gases has internal energy = $\displaystyle \frac {1}{2}\, RT$
0%
Translational degree of freedom of all ideal gases is three
0%
Translational kinetic energy of one mole of all ideal gases is $\displaystyle \frac{3}{2} RT$

Explanation

Translational kinetic energy of an ideal gas in one degree of freedom is

$\dfrac{1}{2}nRT$ , where

$n$ is the number of moles of ideal gas.

Translational kinetic energy of an ideal gas in three degrees of freedom is

$\dfrac{3}{2}nRT$ ,. Thus gases at same temperature may have different kinetic energies if the moles of the gases are different.

$\Delta C_p$ for change

$N_2(g)+3H_2 (g)= 2N\! H_3(g)$ is:

0%
$C_{pNH_3}-(C_{pN_2})$
0%
$2C_{pNH_3}-(C_{pN_2}+3C_{pH_2})$
0%
$2C_{pNH_3}-(C_{pH_2})$
0%
$2C_{pNH_3}+(C_{pN_2}+3C_{pH_2})$

Explanation

$\Delta C_p$ represents the change in the heat capacity at constant pressure. It is equal to the difference in the heat capacities of the products and the reactants.

$\Delta C_p = C_p \: of \: products - C_p \: of \: reactants$ .

$\Delta C_p$ for change

$N_2(g)+3H_2 (g)= 2N\! H_3(g)$ is given by the following expression.

$\Delta C_p=2C_{pNH_3}-(C_{pN_2}+3C_{pH_2})$

The poisson's ratio for

$O_2$ is

$1.4$ . Which of the following are correct for

$O_2$ ?

0%
$c_{ v } = 5 \ cal \ K^{-1} \ mol^{-1}$
0%
$C_v = 0.156\ cal\ K^{-1}\ g^{-1}$
0%
$c_p = \dfrac { R\gamma }{ \gamma - 1 }$
0%
$c_v = \dfrac { R }{ (\gamma - 1) }$

Explanation

$\dfrac { c_p }{ c_v} = 1.4 = \gamma$

$c_p - c_v = R$

$\therefore$

$c_v = \dfrac { R }{ \gamma - 1 }$ and similarly

$c_p - \dfrac { c_p }{ \gamma } = R$

$\therefore$

$c_v = \dfrac { 2 }{ 0.4 } = 5\ cal\ K^{-1}\ mol^{-1}$

$\therefore$

$c_p = \dfrac { \gamma \cdot R }{ (\gamma - 1) }$
Also,

$C_v = \dfrac { 2 }{ 0.4\times 32 } = 0.156\ cal\ K^{-1}\ g^{-1}$

When x amount of heat is given to a gas at constant pressure, it performs

$\displaystyle \frac{x}{3}$ amount of work. The average number of degrees of freedom per molecule of the gas is-

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

Explanation

$\displaystyle \frac{W}{Q}=\frac{P\Delta V}{nC_{P}\Delta T}=\frac{nR\Delta T}{nC_{P}\Delta T}=\frac{x/3}{x}$ (standard result)

$\displaystyle \Rightarrow C_{P}=3R=\left ( \frac{f}{2}+1 \right )R\Rightarrow f=4$

At ordinary temperatures, the molecules of a diatomic gas have only translational and rotational kinetic energies. At high temperatures, they may also have vibrational energy. As a result of this compared to lower temperatures, a diatomic gas at higher temperatures will have-

0%
lower molar heat capacity
0%
higher molar heat capacity
0%
lower isothermal compressibility
0%
higher isothermal compressibility

Which of the following expands most on the rise of an equal amount of temperature?

0%
Liquid
0%
Solid
0%
Gas
0%
Cannot say

A sample of an ideal gas is contained in a cylinder. The volume of the gas is suddenly decreased. A student makes the following statements to explain the change in pressure of the gas.
I. The average kinetic energy of the gas atoms increases.
II. The atoms of the gas hit the walls of the cylinder more frequently.
III. Temperature of the gas remains unchanged.
Which of these statements is true?

0%
I and II only
0%
I and III only
0%
II and III only
0%
I, II and III

STATEMENT-1 : According to kinetic theory of gases the internal energy of a given sample of an ideal gas is only kinetic.
STATEMENT-2 : The ideal gas molecules exert force on each other only when they collide.

0%
STATEMENT-1 is True, STATEMENT-2 is True; STATEMENT-2 is a correct explanation for STATEMENT-1
0%
STATEMENT-1 is True, STATEMENT-2 is True; STATEMENT-2 is NOT a correct explanation for STATEMENT-1
0%
STATEMENT-1 is True, STATEMENT-2 is False
0%
STATEMENT-1 is False, STATEMENT-2 is True

When the temperature is increased:

0%
speed with which the molecules vibrate increases
0%
speed with which the molecules vibrate decreases
0%
there is no change in the speed of the molecules
0%
the speed of the molecules remains constant

Explanation

$\textbf{Explanation}$

$\bullet$ Kinetic energy of particle can be given by the formula:

$K = \dfrac{3}{2} K_b\ T$ , where $K_b$ is Boltzmann’s constant and T is temperature.

$\bullet$ As temperature increases, it will results into increase of kinetic energy. Electrons or atoms in gaseous states are bound with either energy states or covalent bonds which results into increment of vibrational energy.

So, speed with which the molecules vibrate increases.

Hence, option A is the correct answer.

Average value of poisson`s ratio for a mixture of two mol of each gas A and B is

$1.66$ , then

0%
gases are mono-atomic
0%
gases are diatomic
0%
average molar heat capacity at constant volume is $4 cal$
0%
average molar heat capacity at constant V is $3 cal$ .

Explanation

$(a,c)$

$\Upsilon$ for mono-atomic is

$1.66$ .

$\therefore$

$C_V = \frac { 3R }{ 2 }, C_P = \frac { 5 }{ 2 }R$

$C_{ VM } = \frac { 2\times \frac { 3R }{ 2 } + 2\times \frac { 3R }{ 2 } }{ 4 }$

$\therefore$

$C_V = 3 cal$

$C_{ PM } = \frac { 2\times \frac { 5R }{ 2 } + 2\times \frac { 5R }{ 2 } }{ 4 }$

$\therefore$

$C_P = 5 cal$

One mole of a diatomic gas undergoes a process

$P=\frac {P_0}{1+\left (\frac {V}{V_0}\right )^3}$ , where

$P_0, V_0$ are constants. The translational kinetic energy of the gas when

$V = V_0$ is given by

0%
$\frac {5P_0V_0}{4}$
0%
$\frac {3P_0V_0}{4}$
0%
$\frac {3P_0V_0}{2}$
0%
$\frac {5P_0V_0}{2}$

Explanation

$\displaystyle P=\dfrac {P_0}{1+\left (\dfrac {V}{V_0}\right )^3}=\dfrac {P_0}{2}\Rightarrow T=\dfrac {P_0V_0}{2R}$

$\therefore$ Translational kinetic energy is equal to $\dfrac {3}{2}RT=\dfrac {3R}{2}\dfrac {P_0V_0}{2R}=\dfrac {3P_0V_0}{4}$

0%
250
0%
500
0%
-250
0%
-500

A gas behaves more closely like an ideal gas at

0%
low pressure and low temperature
0%
low pressure and high temperature
0%
at all pressure and temperature
0%
none of these

If T represents the absolute temperature of an ideal gas, the volume coefficient of thermal expansion at constant pressure is proportional to:

0%
T
0%
T $^2$
0%
1/T
0%
1/T $^2$

Explanation

Coefficient of thermal expansion at constant pressure is:

$\implies { \alpha }_{ V }=\dfrac { 1 }{ V } \dfrac { dV }{ dT }$

$PV=nRT$

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

${ \alpha }_{ V }=\dfrac { P }{ nRT } \dfrac { dV }{ dT }$

${ \alpha }_{ V }\rightarrow \dfrac { 1 }{ T }$

The poisson`s ratio for

$O_2$ is

$1.4$ . Which of the following are correct for

$O_2$ ?

0%
$C_{ VM } = 5 cal$
0%
$C_V = 0.156 cal$
0%
$C_P = \frac { R\Upsilon }{ \Upsilon - 1 }$
0%
$C_V = \frac { R }{ (\Upsilon - 1) }$

Explanation

$(a,b,c,d)$

$\frac { C_P }{ C_V } = 1.4 = \Upsilon$

$C_P - C_V = R$

$\therefore$

$C_V = \frac { R }{ \Upsilon - 1 }$ and similarly

$C_P - \frac { C_P }{ \Upsilon } = R$

$\therefore$

$C_V = \frac { 2 }{ 0.4 } = 5 cal$

$\therefore$

$C_P = \frac { \Upsilon \cdot R }{ (\Upsilon - 1) }$
Also,

$C_V = \frac { 2 }{ 0.4\times 32 } = 0.156 cal$

A certain mass of an ideal gas undergoes a reversible isothermal compression. Its molecules, compared with the initial state, will then have the same
(i) root mean square velocity
(ii) mean momentum
(iii) mean kinetic energy

0%
(i), (ii), (iii)
0%
(i), (ii)
0%
(ii), (iii)
0%
(i)

Explanation

Given process is isothermal i.e,

$T$ is constant so, root mean velocity

$=\sqrt { \dfrac { 3KT }{ M } }$ is constant mean momentum is always zero mean kinetic energy is

$\dfrac{1}{2}MV^2_{av}$ where

$\Rightarrow V_{av}=\sqrt { \dfrac { 8KT }{ \pi M } }$ is constant at constant

$T.$

Hence,

$(i),(ii),(iii)$ are correct answer.

Kinetic energy of the particles:

0%
is the energy of the bulk matter
0%
decreases with increase in temperature
0%
increases with increase in temperature
0%
temperature has no effect on kinetic energy

The number of air molecules in a ( 5m x 5m x 4m ) room at standard temperature and pressure is of the order of :

0%
$6\, \times\, 10^{23}$
0%
$3\, \times\, 10^{24}$
0%
$3\, \times\, 10^{27}$
0%
$6\, \times\, 10^{30}$

Explanation

Volume is

$100m^3$

$PV=NKT$

$10^5\times 100=N\times1.38\times10^{-23}\times 273$

$N=2.69\times10^{27}=3\times10^{27}$

The mean kinetic energy of a gas molecule is proportional to

0%
$\displaystyle \sqrt { T }$
0%
$\displaystyle { T }^{ 3 }$
0%
$\displaystyle T$
0%
None of the above

Which of the following statement is incorrect for gases?

0%
Mass of gas cannot be determined by weighing the container in which it is enclosed.
0%
Gases do not have a definite shape and volume.
0%
The volume of a gas is equal to the volume of the container confining the gas.
0%
Confined gas exerts uniform pressure on the walls of its container in all directions.

When the temperature is increased from 0

$^o$ C to 273

$^o$ C, in what ratio the average kinetic energy of molecules changes?

0%
1
0%
5
0%
4
0%
2

Explanation

Average K.E.

$= \displaystyle \frac{3}{2}$ RT
At 0

$^o$ C, average K.E.

$= \displaystyle \frac{3}{2} \times R \times 273$

$[T = (0 + 273) K]$
At 273

$^o$ C,
average K.E.

$= \displaystyle \frac{3}{2} \times R \times (273 + 273)$

$= \displaystyle \frac{3}{2} \times R \times 2 \times 273$

$\therefore$ Ratio = 2

The internal energy of a gas:

0%
is the sum total of kinetic and potential energies.
0%
is the total transitional kinetic energies
0%
is the total kinetic energy of randomly moving molecules.
0%
is the total kinetic energy of gas molecules

A quantity of air

$(\gamma = 1.4)$ at 27

$^o$ C is compressed suddenly, the temperature of the air system will

0%
fall
0%
rise
0%
remain unchanged
0%
first rise and then fall

The variation of the product of pressure and volume

$PV$ with volume

$V$ of a fixed mass of an ideal gas at constant temperature is graphically represented by the curve

Which one of the following represents the correct arrangement in the increasing order of forces of attraction between their particles?

0%
Water, air, oxygen
0%
Air, sugar, oil
0%
Oxygen, water, sugar
0%
None of these

Explanation

Gas molecules are at maximum distance from each other and have minimum force of attraction between them. Liquids have more force of attraction than gases while solids have maximum force of attraction.

So the given order is

$oxygen,water \ and \ sugar.$

Which of the following is the correct increasing order of the force of attraction between their particles?

0%
Gold < water < Honey < Air
0%
Air< Gold < Water < Honey
0%
Air < Water < Honey < Gold
0%
Water < Honey < Air < Gold

Explanation

Air < Water < Honey < Gold

it is because the force of attraction increases in the order Gas < Liquid < Solid

and Air is gas

Water is liquid and Honey is a viscous liquid and Gold is solid

Hence, te correct option is

$\text{C}$

For a given gas, which of the following relationships is correct at a given temp?

0%
$u_{rms} > u_{av} > u_{mp}$
0%
$u_{rms} < u_{av} < u_{mp}$
0%
$u_{rms} > u_{av} < u_{mp}$
0%
$u_{rms} < u_{av} > u_{mp}$

Explanation

$u_{rms} =$ Root mean square velocity

$u_{ar}=$ Average velocity

$u_{mp}$ = Most probable velocity

$u_{mp}:u_{ar}:u_{rms}= 1: 1.128: 1224$

$\therefore u_{rms} > u_{ar} > u_{mp}$

The quantity of heat required to heat 1 mole of a monoatomic gas through one degree K at constant pressure is:

0%
3.5 R
0%
2.5 R
0%
1.5 R
0%
none of these

Explanation

Quantity of hear required

$= n.C_p. \Delta T$
Here,

$n= 1, \Delta T = K$
For monoatomic gases,

$C_p=C_V+R$

$=1.5 R+R =2.5R$ .

$\therefore$ Heat required

$= C_p= 2.5 R$

The volume of mole of a prefect gas at NTP is ______.

0%
22.4 litres
0%
2.24 litres
0%
100 litres
0%
none of these

Explanation

At NTP is normal temperature and pressure.

$\Rightarrow$ Pressure

$=1\;atm\;;T=293k\;;R=0.08205\dfrac{L\;atm}{mol}$

$\Rightarrow V=\dfrac{RT}{P}$

$\Rightarrow V=\dfrac{(0.08205)293}{1\;atm}$

$\Rightarrow V_{molar}=22.4\;L$

Hence, the answer is

$22.4\;L.$

The volume of a perfect gas at NTP is

0%
22.4 litres
0%
2.24 litres
0%
100 litres
0%
None of these

Explanation

At NTP

Pressure

$=1$ atm

Temperature

$=293k$

According to ideal gas equation

$PU=nRT$

$\Rightarrow V=\dfrac{1\;mol\;(0.08205)293}{1\;atm}$

$\Rightarrow (For\;n=1)\;V=22.4\;L$

Hence, the answer is

$22.4\;L.$

When we cool a gas below its condensation point, the KE of its molecules will

0%
increase
0%
decrease
0%
remains the same
0%
first increase then decrease

When an inflated tyre bursts, the air escaping out __________.

0%
will get heated up
0%
will be cooled
0%
will not undergo any change in its temperature
0%
will be liquefied

Explanation

Bursting of a tyre is an adiabatic expansive i.e,

$dQ=0$

$\&$

$dV=+ve$

So, According to first law of thermodynamic

$dQ=dV+PdV$

$\Rightarrow -PdV=dV=CvdT$

$P$ is

$+ve$

$dV$ is

$+V$ thus

$dV=-ve$

So,

$dT$ is

$-ve$ means decrease in temperature.

Hence, the answer is will be cooled.

The temperature to which a gas must be cooled before it can't be liquefied by pressure alone is called its

0%
Critical temperature
0%
Saturation point
0%
Boiling point
0%
Freezing point

Explanation

$Answer:-$ A

In this P-V curve the isotherm having

$T_c$ is that critical temperature and corresponding to that there is critical pressure

$P_c$ .

Critical temperature is the temperature of a gas in its critical state, above which it cannot be liquefied by pressure alone.

Critical pressure is the pressure of a gas or vapour in its critical state.

The molecular kinetic energy of a liquid is:

0%
more than molecular kinetic energy of solids
0%
more than molecular kinetic energy of gases
0%
less than molecular kinetic energy of solids
0%
dependent on the matter

When we cool a gas below its condensation point, the K.E. of its molecules:

0%
increases
0%
decreases
0%
remains the same
0%
first decreases then increases

Real gases obey ideal gas laws more closely at:

0%
low pressure and low temperature
0%
low pressure and high temperature
0%
high pressure and low temperature
0%
high pressure and high temperature

Two gases

$A$ and

$B$ having the same temperature

$T$ , same pressure

$P$ and same volume

$V$ are mixed. If the mixture is at the same volume

$V$ , the pressure of the mixture is

0%
$P$
0%
${P}/{2}$
0%
$2P$
0%
$4P$

Explanation

From Ideal Gas Law,

$PV=nRT$

$\implies$ Moles of a gas=

$n=\dfrac{PV}{RT}$

When the mixture is created, the total number of moles is equal to the sum of that of individual gases.

Hence

$N=n_1+n_2$

$\implies \dfrac{P'V}{RT}=\dfrac{PV}{RT}+\dfrac{PV}{RT}$

$\implies P'=2P$

Which of the following is an assumption of Kinetic theory of matter?

0%
Molecules are in a state of continuous motion and possess kinetic energy.
0%
The kinetic energy of molecules increases with increase in temperature.
0%
The molecules of matter always attract each other due to forces of cohesion and adhesion.
0%
All of the above

The gases carbon-monoxide

$(CO)$ and nitrogen at the same temperature have kinetic energies

${E}_{1}$ and

${E}_{2}$ respectively. Then

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${E}_{1}=E_2$
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${E}_{1}> {E}_{2}$
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${E}_{1}< {E}_{2}$
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${E}_{1}$ and ${E}_{2}$ cannot be compared

Explanation

The gases carbon monoxide (

$CO$ ) and nitrogen (

${N}_{2}$ ) are diatomic, so both have equal kinetic energy

$\cfrac{5}{2}kT$ , ie.,

${E}_{1}={E}_{2}$

CBSE Questions for Class 11 Engineering Physics Kinetic Theory Quiz 7 - 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 7 - MCQExams.com

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Practice Class 11 Engineering Physics Quiz Questions and Answers

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