MCQ Questions for Class 11 Engineering Physics Kinetic Theory Quiz 13

In Maxwell's speed distribution curve, for

N2N2K will be?

0%
$300$ m/sec
0%
$606$ m/sec
0%
$920$ m/sec
0%
Zero

Explanation

Magnitude of relative velocity,

$|V_{rev}|=\sqrt{V^2+V^2-2(V)(V)\cos\theta}=2V\left|\sin\dfrac{\theta}{2}\right|$

Average of Relative velocity

$(V_{rev})=\dfrac{\displaystyle\int^{\pi}_02V\left|\sin\dfrac{\theta}{2}\right|d\theta}{\displaystyle\int^{\pi}_0d\theta}=\dfrac{4V}{\pi}$

$(V_{rev})=\dfrac{4}{\pi}V_{average}=\dfrac{4}{\pi}\sqrt{\dfrac{8RT}{\pi m_0}}=\dfrac{4}{\pi}\sqrt{\dfrac{8\times 8.3\times 300}{3.14\times 28\times 10^{-3}}}=606$ m/sec.

1300769_1631972_ans_640e2a071c0f4c3799459e9a7e477fe4.png

For which of the following ideal gas

${C}_{V,m}$ is independent of temperature?

0%
$He$
0%
${H}_{2}$
0%
$CO$
0%
${SO}_{2}$

Explanation

${C}_{v,m}$ molar heat capacity at constant volume.

It is heat required to increase the temperature of

$1$ mole of the gas by

$1K$ .

But for inert gases, it is independent of temperature.

Option A is correct.

At constant pressure, the heat of formation of a compound is not dependent on temperature, when:

0%
$\Delta C_p = 0$
0%
$\Delta C_v = 0$
0%
$\Delta C_p > 0$
0%
$\Delta C_p < 0$

Explanation

At constant pressure, the enthalpy of element or compound changes with temperature according to the formula:

$\Delta H=\Delta C_p T$

So, if

$\Delta C_p=0$ then

$\Delta H=0$

Thus, the heat of formation of the compound becomes independent of temperature.

Pressure of an ideal gas is increased by keeping temperature constant. The kinetic energy of molecules.

0%
Decreases
0%
Increases
0%
Remains same
0%
Increases or decreases depending on the nature of gas

A quantity of

$10$ g of a gas at

$1$ atm pressure is cooled from

$273^o$ C to

$273$ K keeping its volume constant, the final pressure of the gas will be?

0%
$273$ atm
0%
$0.5$ atm
0%
$0.2$ atm
0%
$0.1$ atm

Explanation

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

$\dfrac{1}{546}=\dfrac{P_2}{273}$

$P_2=\dfrac{273}{546}$

$P_2=\dfrac{1}{2}=0.5$ atm.

Answer option B.

If the temperature of

$3$ moles of helium gas is increased by

$2\ K$ , then the change in the internal energy of helium gas is :

0%
$70.0\ J$
0%
$68.2\ J$
0%
$74.8\ J$
0%
$78.2\ J$

In the case of solid, number of degrees of freedom is :

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

The average kinetic energy of a gas molecule at

$27^oC$ is

$6.21\times 10^{-21}\ J$ . Its average kinetic energy at

$227^oC$ will be :

0%
$52.2\times 10^{-21}\ J$
0%
$5.22\times 10^{-21}\ J$
0%
$10.35\times 10^{-21}\ J$
0%
$11.35\times 10^{-21}\ J$

The curve in figure represents potential energy(U) in between two atoms in a diatomic molecule as a function of distance 'x' between atoms. The atoms are:

0%
Attracted when x lies between 'A' and 'B' and repelled when x lies between B and C
0%
Attracted when x lies between B and C and repelled when x lies between A and B
0%
Attracted when they reach B
0%
Repelled when they reach B

Reducing the pressure from

$1.0$ atm to

$0.5$ atm would change the number of molecules in one mole of ammonia to?

0%
$75\%$ of initial value
0%
$50\%$ of initial value
0%
$25\%$ of initial value
0%
None of these

Explanation

Pressure is decreased from

$1$ atm to

$0.5$ atm i.e., pressure increased or decreased doesn't affect the mole of ammonia
i.e., the no. of molecules in

$1$ mole of ammonia won't change

$=6.022\times 10^{23}$ .

At what temperature the

$KE$ of gas molecule is half that of its volue at

$27^oC$ ?

0%
$13.5^oC$
0%
$150^oC$
0%
$150\ K$
0%
$-123\ K$

The average translation kinetic energy of

$O_2$ (molar mass

$32$ ) molecules at a particular temperature is

$0.048\ eV$ . The translational kinetic energy of

$N_2$ (molar mass

$28$ ) molecules in

$eV$ at the temperature is :

0%
$0.0015$
0%
$0.003$
0%
$0.048$
0%
$0.768$

The quantity

$\dfrac{PV}{kT}$ represents

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

The ideal gas equation is given as:

$PV=nRT$

The Boltzmann constant

$k$ is represented as:

$k=\dfrac{R}{N}$

Now, substitute the value of the Boltzmann constant :

$PV=nkNT$

$\therefore \dfrac{PV}{kT}=nN$

It is the number of molecules of the particular sample. SO, option

$D$ is correct.

Which of the following parameters is the same for molecules of all gases at a given temperature?

0%
Mass
0%
Speed
0%
Momentum
0%
Kinetic energy

Explanation

At a particular value of the temperature, the average kinetic energy of all the molecules is considered to be constant as temperature is the only factor that determines the kinetic energy of the molecules for the particular sample of gas.

Thus, option

$D$ is correct.

The heat capacity of air is

$20\ J/K/$ mol. The amount of heat, required to heat the room through

$1^o$ , assuming the amount of air in the room to be

$29\ kg$ , is

0%
$20\ kJ$
0%
$-20\ J$
0%
$200\ J$
0%
$-200\ kJ$

Explanation

Dry air contains

$21 \%$ oxygen and

$78 \%$ nitrogen.

Thus molar mass of air = 29 g

The amount of heat required =

$n(moles) \times C \times \triangle T = \dfrac{29000}{29} \times 20 \times 1 = 20k J$

In Fig. here shows four paths traversed by a gas on a P-V diagram. If

$\Delta U_1, \Delta U_2, \Delta U_3$ and

$\Delta U_4$ are the change in internal energies in their respective path, then choose the incorrect relation:

0%
$\Delta U_1 = \Delta U_2 = \Delta U_4 = \Delta U_3$
0%
$\Delta U_1 + \Delta U_2 = \Delta U_3 + \Delta U_4$
0%
$\Delta U_1 > \Delta U_2 > \Delta U_3 > \Delta U_4$
0%
$\Delta U_1 < \Delta U_2 < \Delta U_3 < \Delta U_4$

The energy density U / V of an ideal monoatomic gas is related to its pressure P as

0%
$\dfrac {U}{V} = 3P$
0%
$\dfrac {U}{V} = \dfrac {3}{2} P$
0%
$\dfrac {U}{V} = \dfrac {P}{3}$
0%
$\dfrac {U}{V} = \dfrac {5}{2} P$

Explanation

$PV=RT$

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

$U$ is the energy of gas

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

$\dfrac{U}{V}= \dfrac{3P}{2}$

$\dfrac{U}{V}$ is the energy density

The ratio of average translational kinetic energy to rotational kinetic energy of a diatomic molecule at temperature

$T$ is :

0%
$3$
0%
$7/5$
0%
$5/3$
0%
$3/2$

Find the approx. number of molecules contained in a Vessel of volume 7 litres at

$0^0 C\ at\ 1.3 \times 10^5$ pascal:

0%
$2.4 \times 10^{23}$
0%
$3.0 \times 10^{23}$
0%
$6.0 \times 10^{23}$
0%
$4.8 \times 10^{23}$

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 hear added
0%
must always decrease
0%
will remain the same if change in internal energy equals the heat added

Select the correct statements about ideal gas:

0%
Molecules of a gas are in random motion colliding against one another and with the walls of the container
0%
The gas is not isotropic and the constant (1/3) in equation $p = \frac{1}{3}pv^2$ r. m. s. is result of this property
0%
The time during which a collision lasts is negligible compared to the time of free path between collisions
0%
There is no force of interaction between molecules among themselves or between molecules and the wall except during collision.

An ideal gas undergoes the cyclic process shown in a graph below

0%
$T_1 = T_2$
0%
$T_1 > T_2$
0%
$V_aV_c = V_bV_d$
0%
$V_aV_b = V_cV_d$

Explanation

$\text{For adiabatic process 'bc'}$

$T_1V{^{\gamma-1}_B}=T_2V{^{\gamma-1}_C}......(i)$

$\text{For adiabatic process 'da'}$

$T_2V{^{\gamma-1}_d}=T_1V{^{\gamma-1}_a}......(ii)$

Multiplying (i) and (ii)

$\Rightarrow T_1T_2(V_b V_d)^\gamma-1=T_1T_2(V_aV_c)^\gamma-1$

$\Rightarrow V_bV_d=V_aV_c$

Hence option 'B' and 'D' are correct

A lead bullet just melts when stopped by an obstacle. Assuming that 25% of heat is absorbed by the obstacle, find the minimum velocity of the bullet if its initial temperature is

$27^{0}C$ (melting point of lead

$327^{0}C$ ; specific heat of lead

$= 0.03 \space cal/g^{0}C$ ; latent heat of fusion of lead = 6 cal/g and J = 4.2 J/cal)

0%
450 m/s
0%
398 m/s
0%
420 m/s
0%
410 m/s

Explanation

c. If mass of the bullet is m grams, heat absorbed by it to raise its temperature from

$27^{0}C$ to

$327^{0}C$

$mL = m \times 6 = 6 m\ cal$

So, the total heat required by the bullet

$Q_1 = (9m + 6m) = 15 m cal = (15m \times 4.2) J$
(as 1 cal = 4.2 J)
Now when the bullet is stopped by the obstacle, loss in its mechanical energy.

$ME = \dfrac {1} {2} (m \times 10^{-3})v^{2} J$ (as m gram =

$m \times 10^{-3}$ kg)

As 25% of the energy is absorbed by the obstacle, the energy absorbed by the bullet

$Q_2 = \dfrac{75} {100} \times \dfrac {1} {2} mv^{2} \times 10^{-3} = \dfrac {3} {8} mv^{2} \times 10^{-3} J$

Now the bullet will melt if

$Q_2 \geq Q_1$

$\dfrac {3} {8} mv^{2} \times 10^{-3} \geq 15 m \times 4.2$

$v \geq \sqrt {(4 \times 4.2)} \times 10^{2}$

$v_{min} = 410 m/s$

Postulate of kinetic theory is

0%
Atom is indivisible
0%
Gases combine in a simple ratio
0%
There is no influence of gravity on the molecules of a gas
0%
None of the above

If Cv = 4.96 cal / mole K, then increase in internal energy when temperature of 2 moles of this gas is increased from 340 K to 342 K

0%
$27.80$ cal
0%
$19.84$ cal
0%
$13.90$ cal
0%
$9.92$ cal

Explanation

Change in internal energy is always equal to the heat supplied at constant volume.

$\Delta U = \left ( \Delta Q \right )_{v}= \mu C_{v}\Delta T.$

So,

$\Delta U=2\times 4.96\times (342-340)=19.84 cal$

The temperature of an ideal gas is kept constnat as it expands. The gas does external work. During this process, the internal energy of the gas

0%
Decreases
0%
Increases
0%
Remains constant
0%
Depend on the molecular motion

Molecules of a gas behave like

0%
Inelastic rigid sphere
0%
Perfectly elastic non-rigid sphere
0%
Perfectly elastic rigid sphere
0%
Inelastic non-rigid sphere

Heat is not being exchanged in a body. If its internal energy is increased then

0%
Its temperature will increase
0%
Its temperature will decrease
0%
Its temperature will remain constant
0%
None of these

The internal energy of an ideal gas depends upon

0%
Specific volume
0%
Pressure
0%
Temperature
0%
Density

The degrees of freedom of a stationary rigid body about its axis will be

0%
One
0%
Two
0%
Three
0%
Four

The pressure exerted by the gas on the walls of the container because

0%
It loses kinetic energy
0%
It sticks with the walls
0%
On collision with the walls there is a change in momentum
0%
It is increased towards the walls

The number of translational degrees of freedom for a diatomic gas is

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

A monoatomic gas molecules has

0%
Three degrees of freedom
0%
Four degrees of freedom
0%
Five degrees of freedom
0%
Six degrees of freedom

The degrees of freedom of a triatomic gas is

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

Explanation

The general expression 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,

$DOF=(3\times 3)−3$

$DOF=9−3$

$DOF=6$

At a given volume and temperature , the pressure of a gas

0%
Varies inversely as its mass
0%
Varies inversely as the square of its mass
0%
Varies linearly as its mass
0%
Is independent of its mass

A diatomic molecules has how many degrees of freedom

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

In an ideal gas, the molecules possess

0%
only kinetic energy
0%
both kinetic energy and potential energy
0%
only potential energy
0%
neither kinetic energy nor potential energy

Explanation

In an ideal gas, the molecules only have

$\textbf{ kinetic energy}$

The mean kinetic energy of a gas at

$300\,K$ is

$100J$ . The mean energy of the gas at

$450\,K$ is equal to

0%
$100\,J$
0%
$3000\,J$
0%
$450\,J$
0%
$150\,J$

Mean kinetic energy per degree of freedom of gas molecules is

0%
$\dfrac{3}{2} kT$
0%
$kT$
0%
$\dfrac{1}{2}kT$
0%
$\dfrac{3}{2} RT$

On colliding in a closed container the gas molecules

0%
Transfer momentum to the walls
0%
Momentum becomes zero
0%
Move in opposite directions
0%
Perform Brownian motion

A thin walled hollow cylinder is rolling down an incline , without slipping. At any instant, without slipping. At any instant the ratio "Rotational K.E : Translational K.E : Total K.E" is

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

According to the law of equipartition of energy, the energy associated with each degree of freedom is:

0%
$\frac{1}{3}K_{B}T$
0%
$\frac{1}{2}K_{B}T$
0%
$K_{B}T$
0%
$\frac{3}{2}K_{B}T$

Explanation

According to the law of equipartition of energy, the energy associated with each degree of freedom is

$\frac{1}{2}K_{B}T$

Hence, Option "B" is the correct answer.

A 90 cm long barometer tube contains some air above the mercury. The reading is 74.5 cm when the true pressure is 76 cm at the temperature

$15^{0} C$ . If the reading is observed to be 75.8 cm on a day when the temperature is

$5^{0}C$ , then the true pressure is:

0%
77.38 cm of Hg
0%
75.8 cm of Hg
0%
74 cm of Hg
0%
80 cm of Hg

Explanation

Pressure = Actual pressure - Observed pressure

Volume = A(Barometer length - observed length)

$\dfrac { { P }_{ 1 }{ V }_{ 1 } }{ { T }_{ 1 } } =\dfrac { { P }_{ 2 }{ V }_{ 2 } }{ { T }_{ 2 } } \\ \Rightarrow \dfrac { (76-74.5)(90-74.5) }{ 288 } =\dfrac { (P-75.8)(90-75.8) }{ 278 } \\ \Rightarrow P=77.38 \ \ cm \ \ of \ Hg$

Two rigid boxes containing different ideal gases are placed on a table. Box A contains one mole of

$N_{2}$ at temperature

$T_{0}$ , while box B contains one mole of

$H_{2}$ at temperature 7/3

$T_{0}$ . The boxes are then put into thermal contact with each other and heat flows between them until the gases reach a common final temperature [Ignore the heat capacity of boxes]. Then the final temperature of the gases

$T_{f}$ in terms of

$T_{0}$ is

0%
$T_{f}=\dfrac{5}{2}T_o$
0%
$T_{f}=\dfrac{3}{7}T_o$
0%
$T_{f}=\dfrac{5}{3}T_o$
0%
$T_{f}=\dfrac{3}{2}T_o$

Explanation

According to kinetic theory of gases, the KE of an ideal gas molecule at temp

$T$ is given by

$\dfrac{1}{2}fk_BT$ where f is the no of degrees of freedom,

$k_B$ is the Stefan Boltzman constant and T is the temp in Kelvin.
Both Nitrogen and Hydrogen are diatomic and the no of degrees of freedom

$f=5$ .

Total KE is

$KE=\dfrac{1}{2}fn_1N_Ak_BT_1 + \dfrac{1}{2}fn_2N_Ak_BT_2=\dfrac{1}{2}fN_Ak_B(n_1T_1+n_2T_2)$ ....(1) where

$N_A$ is the Avogadro's number.

$T_1=T_0$ ,

$T_2=\frac{7}{3}T_0$

$n_1=n_2=n$
Let the final temp be

$T_f$ after both gases are mixed.

$KE= \dfrac{1}{2}(n_1+n_2)fN_Ak_BT$ ...(2)

Equating (1) and (2) since total energy is conserved,

$\dfrac{1}{2}(n_1+n_2)fN_Ak_BT_f = \dfrac{1}{2}fN_Ak_B(n_1T_1+n_2T_2)$

$\therefore T_f=\dfrac{n_1T_1+n_2T_2}{n_1+n_2}=\dfrac{1}{2}\times (T_1+ T_2)=\dfrac{1}{2}\times \dfrac{10}{3}T_0=\dfrac{5}{3}T_0$

Internal energy of

$n_{1}$ moles of

$H_{2}$ at temperature

$T$ is equal to the internal energy of

$n_{2}$ moles of He at temperature

$2T$ . Then the ratio

$\dfrac{n_{1}}{n_{2}}$ is

0%
3/5
0%
2/3
0%
6/5
0%
3/7

Explanation

Internal energy per degree of freedom:

$E=\frac{1}{2}kT$ where k is the Boltzman's constant and T is the temp in Kelvin.
For n_1 moles of H_2

$E_1= n_1 \times f_1\times \frac{1}{2}kT_1$
For n_2 moles of He

$E_2= n_2 \times f_2 \times \frac{1}{2}kT_2$
Given

$E_1=E_2$ and

$T_2=2T_1$

$f_1$ =5 since

$H_2$ is a diatomic molecule and has 5 degrees of freedom where as He is mono atomic and has 3 degrees of freedom.

$\therefore n_1 \times f_1\times \dfrac{1}{2}kT_1= n_2 \times f_2 \times \dfrac{1}{2}kT_2$

$\therefore \dfrac{n_1}{n_2}=\dfrac{f_2T_2}{f_1T_1}=\dfrac {3 \times 2T_1}{5T_1}=\dfrac {6}{5}$

A closed container of volume 0.02 m $^3$ contains a mixture of neon and argon gases at a temperature of 27 $^{0}$ C and at a pressure of $1\times 10^{5}N/m^{2}$ . The total mass of the mixture is 28 g. If the gram molecular weights of neon and argon are 20 and 40 respectively, the masses of the individual gases in the container are respectively(assuming them to be ideal) [R = 8.314 J/mol K]

0%
16 gm, 12 gm
0%
4 gm, 24 gm
0%
6 gm, 22 gm
0%
12 gm, 16 gm

Explanation

$n=\frac { PV }{ RT } \\ n=\frac { 1\times { 10 }^{ 5 }\times 0.02 }{ 8.314\times 300 } \\ n=0.802\\ { n }_{ 1 }{ +n }_{ 2 }=0.802\\ { 20n }_{ 1 }{ +20n }_{ 2 }=16\quad \left( Multiplying\quad by\quad 20 \right) \\ { w }_{ 1 }{ +w }_{ 2 }=28\\ { n }_{ 1 }{ M }_{ neon }{ +n }_{ 2 }{ M }_{ argon }=28\\ 20{ n }_{ 1 }{ +40n }_{ 2 }=28\\ Solving\quad to\quad two\quad equations\quad we\quad get,\\ 20{ n }_{ 2 }=12\\ { n }_{ 2 }=0.6\\ \therefore { w }_{ 2 }=0.6\times 40=24gms\\ { w }_{ 1 }=28-24=4gms$

A current passes through a resistor. Let

$K_{1}$ and

$K_{2}$ represent the average kinetic energy of the conduction electrons and the metal ions respectively, then:

0%
$K_{1}$ < $K_{2}$
0%
$K_{1} = K_{2}$
0%
$K_{1}$ > $K_{2}$
0%
Any of these 3 may be correct

A smooth vertical tube with two different cross-sections is open at both ends. They are fitted with pistons of different areas of cross- section and each piston moves within a particular section. One mole of a gas enclosed between the pistons which are tied with non-stretchable threads. The difference in cross-sectional area of pistons

$10\mathrm{c}\mathrm{m}^{2}$ The mass of gas confined between pistons is 5kg. The outside pressure is 1 atmosphere

$=10^{5}N/m^{2}$ . By how many degrees must the gas between pistons be heated to shift the piston by 5 cm?

Given

$R=8.3$ .

0%
$0.9^{\mathrm{o}}\mathrm{C}$
0%
$1.2^{\mathrm{o}}\mathrm{C}$
0%
$1.5^{\mathrm{o}}\mathrm{C}$
0%
$0.5^{\mathrm{o}}\mathrm{C}$

Explanation

Let us consider

$s_1$ and

$s_2$ as the cross-section area of the lower and upper piston respectively.

Let

$P_o$ be the atmospheric pressure and P be the gas pressure.At equilibrium, the downward force and upward forces balances,

i.e.,

$P_o + Ps_1 + mg = Ps_2 + P_{o}s_{1}$ (

$Force = Pressure \times Area$ }

$P = P_{o} + \dfrac{mg}{s_1 - s_2}$ -- (1)

Let

$l_1$ and

$l_2$ be the length of string in lower and upper section.

Initial volume of the gas =

$l_1 s_1 + l_2 s_2$

When the pistons shifts by a distance

$x$ ,

Final volume =

$(l_1 - x)s_1 + (l_2 + x)s_2$

We have

$PV = RT$ . Applying this before heating and after heating the gas,

$P(l_1 s_1 + l_2 s_2) = RT$ -- (2)

$P[(l_1 -x)s_1 + (l_2 + x)s_2] = R (T + \Delta T)$

$P[(l_1 s_1 + l_2 s_2)+ (s_2 - s_1)x]= R (T + \Delta T)$

$[P(l_1 s_1 + l_2 s_2)+ P(s_2 - s_1)x]= R (T + \Delta T)$

$P(l_1 s_1 + l_2 s_2)+ P \Delta S x= R (T + \Delta T)$

Substituting for

$P(l_1 + l_2)$ from equation 2,

$RT+ P \Delta S x= R (T + \Delta T)$

Substituting for

$P$ from equation 1,

$RT+ P_{o} + \dfrac{mg}{s_1 - s_2} \Delta S x= R (T + \Delta T)$

$\implies \Delta T = \dfrac{1}{R} [P_o + mgx)$

$= \dfrac {1}{8.3} [10^5 \times 5 \times 10^{-2} \times 10 \times 0^{-4} + 5 \times 9.81 \times 5 \times 10^{-2} ]$

$= 0.9$

A horizontal uniform glass tube of 100cm length is sealed at both ends contains 10 cm mercury column in the middle the temperature and pressure of air on either side of mercury column are respectively 31 $^{0}$ C and 76cm of mercury if the air column at one end is kept at 0 $^{0}$ C and the other end at 273 $^{0}$ C the pressure of air which is at 0 $^{0}$ C is (in cm of Hg )

0%
$76$
0%
$88.2$
0%
$102.4$
0%
$122$

Explanation

Let

$x$ be the displacement of the mercury column on the side whose temperature is changed to

$0^{\circ}C$ . New pressure on both sides must be same so that the mercury column remains in equilibrium.

$\dfrac{PV}{T}$ is constant for air column.

$\implies \dfrac{P_{0}(45)}{273+31}=\dfrac{P_1 (45-x)}{273+0}$

and

$\dfrac{P_0 (45)}{273+31}=\dfrac{P_1(45+x)}{273+273}$

Therefore,

$x=15\ cm$ and thus

$P=102.4\ cm\ of\ Hg$

Two identical containers each of volume V $_{0}$ are joined by a small pipe. The containers contain identical gases at temperature T $_{0}$ and pressure P $_{0}$ . One container is heated to temperature 2T $_{0}$ while maintaining the other at the same temperature. The common pressure of the gas is P and n is the number of moles of gas in container at temperature 2T $_{0}$ .

0%
$P=2P_{0}$
0%
$P=\dfrac{4}{3}P_{0}$
0%
$n=\dfrac{2P_{0}V_{0}}{3RT_{0}}$
0%
$n=\dfrac{3P_{0}V_{0}}{2RT_{0}}$

Explanation

Since the total number of moles in the system remains the same before and after heating,

$\dfrac{P_0V_0}{T_0}+\dfrac{P_0V_0}{T_0}=\dfrac{PV_0}{T_0}+\dfrac{PV_0}{2T_0}$

$\implies P=\dfrac{4}{3}P_0$

CBSE Questions for Class 11 Engineering Physics Kinetic Theory Quiz 13 - 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 13 - MCQExams.com

0:0:2

Practice Class 11 Engineering Physics Quiz Questions and Answers

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