Kinetic Theory Of Gases(12th Grade > Physics ) Questions and answers for exam Preparation

12th Grade > Physics

KINETIC THEORY OF GASES MCQs

Total Questions : 24 | Page 1 of 3 pages

Match the following:

1. W_{A B}

I . 0

2. W_{B C}

I I . - 1500 J

3. W_{C A}

I I I . 1000 J

1−I,2−II,3−III
1−II,2−I,3−III
1−III,2−I,3−II
1−III,2−II,3−I

Discuss Question

Answer: Option D. -> 1−III,2−II,3−I
:
D
Take 2-3 minutes and try to get this on your own. Then move on the next video for the explanation.

Question 2. Tea contained in a mug is an example of:

Isolated system
Open system
Closed system
None

Discuss Question

Answer: Option B. -> Open system
:
B
Let's talk about the different kinds of system first.
Open system -matter and energy both can be transferred.
Closed system - Only energy can be transferred.
Isolated - Neither energy nor matter can be transferred.
A mug is a vessel, it allows for the exchange of matter and energy between the system and surroundings. Thus it is an open system.

Question 3. A pressure cooker contains air at 1 atm and

30^{\circ}

C. If the safety value of the cooler blows when the inside pressure

\geq

3 atm, then the maximum temperature of the air, inside the cooker can be

90∘C
636∘C
909∘C
363∘C

Discuss Question

Answer: Option B. -> 636∘C
:
B
Since volume is constant,
Hence

\frac{P_{1}}{P_{2}} = \frac{T_{1}}{T_{2}} \Rightarrow \frac{1}{3} = \frac{(273 + 30)}{T_{2}}

\Rightarrow

T_{2}

= 909 K =

636^{\circ}

Question 4. Find the value of

Δ U

+150 J
-50 J
+50 J
-150 J

Discuss Question

Answer: Option C. -> +50 J
:
C
Take a minute and try to get this on your own. Then move on the next video for the explanation.

Question 5. Two thermally insulated vessels

1

and

2

are filled with air at temperature (

T_{1}

T_{2}

), volume (

V_{1}

V_{2}

) and pressure (

P_{1}

P_{2}

) respectively. If the value joining the two vessels is opened, the temperature inside the vessel at equilibrium will be

T1 + T2
T1 + T22
T1T2(P1V1+P2V2)P1V1T2+P2V2T1)
T1T2(P1V1+P2V2)P1V1T1+P2V2T2)

Discuss Question

Answer: Option C. -> T1T2(P1V1+P2V2)P1V1T2+P2V2T1)
:
C
According to the kinetic theory, the average kinetic energy(KE) per molecule of a gas
=

\frac{3}{2} K T

. Let

n_{1}

and

n_{2}

be the number of moles of air in vessels 1 and 2 respectively.
Before mixing, the total KE of molecules in the two vessels is

E_{1} = \frac{3}{2} n_{1} k T_{1} + \frac{3}{2} n_{2} k T_{2}

\frac{3}{2} k (n_{1} T_{1} + n_{2} T_{2})

After mixing, the total KE of molecules is

E_{2} = \frac{3}{2} (n_{1} + n_{2}) k T

Where T is the temperature when equilibrium is established. Since there is no loss of energy
(Because the vessels are insulated),

E_{2} = E_{1}

\frac{3}{2} (n_{1} + n_{2}) k T = \frac{3}{2} k (n_{1} T_{1} + n_{2} T_{2})

T = \frac{n_{1} T_{1} + n_{2} T_{2}}{n_{1} + n_{2}}

Now

P_{1} V_{1} = n_{1} R T_{1}

and

P_{2} V_{2} = n_{2} R T_{2}

which give

n_{1} = \frac{P_{1} V_{1}}{R T_{1}}

and

n_{2} = \frac{P_{2} V_{2}}{R T_{2}}

Using these in Eq. (1) and simplifying, we get

T = \frac{T_{1} T_{2} (P_{1} V_{1} + P_{2} V_{2})}{(P_{1} V_{1} T_{2} + P_{2} V_{2} T_{1})}

Question 6. A vessel contains

1

mole of

O_{2}

gas (molar mass

32

) at a temperature

T

. The pressure of the gas is

P

. An identical vessel containing one mole of

H e

gas (molar mass

4

) at a temperature

2 T

has a pressure of

Discuss Question

Answer: Option C. -> 2P
:
C
For a gas,

P V = n R T

. Hence

(P)_{O_{2}} = \frac{(1 m o l e) R T}{V}

and

(P)_{H e} = \frac{(1 m o l e) R (2 T)}{V}

∴ \frac{(P)_{H e}}{(P)_{O_{2}}} = 2

(P)_{H e} = 2 (P)_{O_{2}}

Question 7. Pressure versus temperature graphs of an ideal gas are as shown in figure. Choose the wrong statement

Density of gas is increasing in graph (i)
Density of gas is decreasing in graph (ii)
Density of gas is constant in graph (iii)
None of these

Discuss Question

Answer: Option C. -> Density of gas is constant in graph (iii)
:
C
(v)

μ

\frac{P M}{R T}

Density

μ

remains constant when

\frac{P}{T}

or volume remains constant.
In graph (i) Pressure is increasing at constant temperature hence volume is decreasing so density is increasing. Graphs (ii) and (iii) volume is increasing hence, density is decreasing. Not that
volume whould had been constant in case the stright line in graph (iii) had passed through origin

Question 8. A process where the gas is taken from one state to another through infinitely small changes is called

equilibrium process
pseudo-equilibrium
adiabatic process
quasi-static process

Discuss Question

Answer: Option D. -> quasi-static process
:
D
In thermodynamics, a quasi-static process is a thermodynamic process that happens slowly enough for the system to remain in internal equilibrium. Only in such processes can each moment be represented on the P-Vdiagram.

Question 9. The Bat Life: Batman is undercover, and to prevent anyone from recognizing his voice (or Bruce Wayne's) he inhales a very dilute Helium gas (search YouTube for "voice after helium” to see the effects). The helium is stored ina cylindrical chamber as shown.

During a fight one of the steel balls gets dislodged. What will happen to the piston?

It will move down
It will move up
It will remain in the same place
Not enough information, only Batman will know

Discuss Question

Answer: Option B. -> It will move up
:
B
Initial free body diagram -

.
When the system was in equilibrium, the force applied by the steel balls on the piston was offset by the force applied by the gas on the piston.
Just after removing the ball -

.
The instant the ball is dislodged the downward force acting on the piston reduces from 19.6 N

(i . e ., 2 \times 9.8 N) t o 14.7 N (1.5 \times 9.8 N)

. However the pressure of the helium gas has not changed. Hence there is an unbalanced force in the upward direction. This upward force will push the piston up. After some time, the weight of the balls would equal the force exerted by the helium gas and the system would eventually come to rest.

Question 10. Which amongst the following graphs represents the V-T graph shown in the figure?

Discuss Question

Answer: Option B. -> It will move up
:
C
Let us analyse each process separately. BC is an isothermal process, which implies that the P-Cgraph would be represented by a rectangular hyperbola. (Since PV= constant).
This rectangular hyperbola starts at a lower volume

V_{1}

and goes to a higher volume as represented in the figure below -

Process AB is an isochoric process, which in a P-Vgraph would be a vertical line. Also we notice that the temperature drops in this process, whichimplies that the pressure will also drop (since PV=nRT, when Vis constant

P \propto T

). Therefore point Acan be anywhere on the line represented in blue in the figure below.

Process C - A
This process is an isobaric process since it represents a curve where

V \propto T

(this is possible only when pressure is a constant, remember Charles' law?). So the process can be represented by a horizontal line in a P-Vgraph. The point where the line joining process A-Band C-Ameet is the point A.