Multiple choice questions on Kinetic theory of gases for Class 11,JEE and NEET

Assertion and Reason
a. Statement I is true ,statement II is true ,statement II is correct explanation for statement I
b. Statement I is true ,statement II is true ,statement II is not a correct explanation for statement I
c. Statement I is true, Statement II is false
d. Statement I is False, Statement II is True

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Question 17
There are two statement
Statement A. Equal volumes of all gases at the same temperature T and pressure P contain an equal no of Molecules
Statement B,.the no of molecules in one mole of any gas is 6.0255 * 10²².
which one of the following is correct
a. A and B both
b. A only
c B only
d. A and B both are incorrect

Answer

Statement A is Avogadro law. So true
Avogadro no is 6.0255 * 10²³  So second one is true

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Question 18
There are two statement about Ideal gases
Statement A The V_rms of gas molecules depends on the mass of the gas molecule and the temperature
Statement B The V_rms is same for all the gases at the same temperature
which one of the following is correct
a. A and B both
b. A only
c. B only
d. A and B both are incorrect

Answer

v_rms depends on mass and temperature
So A is correct . Ans B is incorrect

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Question 19
STATEMENT 1:The molar specific heat capacity of the ideal gas in isothermal process is infinity
STATEMENT 2:Heat transfer is non zero in the Isothermal process

Answer

The molar specific heat capacity of the ideal gas in isothermal process is infinity as there is no change in temperature
Heat transfer is non zero in the Isothermal process as dU=0 so dQ=dW so dQ not equal to zero

So both the statement is true but statement II is not a correct explanation for statement I

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Question 20
STATEMENT 1:Equation of state for a real gas is (P+a/V²)(V-b)=nRT
STATEMENT 2: Molecular attraction is not negligible and the size of molecules are not negligible in comparison to average separation between them

Answer

Equation of state for a real gas is (P+a/V²)(V-b)=nRT.This statement is true
Molecular attraction is not negligible and the size of molecules are not negligible in comparison to average separation between them. This is also true

Equation of state for a ideal gas is PV=nRT
but Equation of state for a real gas is (P+a/V²)(V-b)=nRT because Molecular attraction is not negligible and the size of molecules are not negligible in comparison to average separation between them

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Question 21
A container has a mixture of 1 mole of oxygen and 2 moles of nitrogen at 330K.The ratio of average rotational kinetic energy per O₂ molecule to that per H₂ molecule is
a. 2:1
b. 1:2
c. 1:1
d. None of these

Answer

Rotational kinetic energy depends only on degree of freedom associated with which is same for both diatomic molecule ie 1:1
So c is correct

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Question 22
1 mole of an ideal gas is contained in a cubical volume V, ABCDEFGH at 300 K One face of the cube (EFGH) is made up of a material which totally absorbs any gas molecule incident on it.
At any given time
(a) the pressure on EFGH would be zero.
(b) the pressure on all the faces will the equal.
(c) the pressure of EFGH would be double the pressure on ABCD.
(d) the pressure on EFGH would be half that on ABCD.

Answer

(d) In the ideal case that we normally consider,each collision transfers twice the magnitude of its normal momentum. On the face EFGH, it transfers only half of that.

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Linked Type Comprehension

According the Maxwell's Speed distribution law
$A(v) = 4 \pi (\frac {M}{2\pi RT})^{3/2} v^2 e^{-Mv^2/2RT}$
Where V is molecular speed
M = molar mass of gas
R = gas constant
T = Temperature
A (v) = Probability distribution function
A (v) dv = fraction of the molecules whose speed lie in the internal of width dv speed center on v

Question 23
What is most probable speed
a. $\sqrt {\frac {2RT}{M}}$
b. $\sqrt {\frac {RT}{M}}$
c. $\sqrt {\frac {3RT}{M}}$
d.$\sqrt {\frac {8RT}{M}}$

Answer

(a)
This is derived from Maxwell equation.Derivation is beyond the scope of the course

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Question 24
When is of these is true with respect to above paragraph where the integral is from 0 to infinity
a. $\int_{0}^{\infty}A (v) dv = 1$
b. $\int_{0}^{\infty}A (v) dv = 2$
c. $\int_{0}^{\infty}A (v) dv = 1/2$
d. $\int_{0}^{\infty}A (v) dv = 0$

Answer

(a)
A(v)dv represent the fraction of molecules...so when all the fractions are considered...this should be equal to 1

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Question 25

let A₁& A₁ are the Area of these two curves which is of the is true
a. A₁ = 1     A₂ = 1          
b. A₁ = 1     A₂ = 2     
c. A₁ = 1,     A₂ = 1.5     
d. None of these

Answer

(a)
A rectangular strip in this curve is represent by =A(v)dv

Area of the curve =∫ A(v)dv
A(v)dv represent the fraction of molecules...so when all the fractions are considered...this should be equal to 1
Area of the curve =1
So

Area under the curve =1 and it is true for all variation.

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Question 26
What is the formula for V_avg
a. $V_{avg} = \int_{0}^{\infty} vA (v) dv $
b. $V_{avg} = \int_{0}^{\infty} A (v) dv $

c. $V_{avg} = \int_{-\infty}^{\infty} vA (v) dv $
d. $V_{avg} = \int_{-\infty}^{\infty} A (v) dv $

Answer

(a)
A(v)dv represent the fraction of molecules
v is the speed.
Now average speed formula is
=(n1v1+n2v2+n3v3...)/(n1+n2+n3)
or
average speed formula =(fraction of molecules have speed v1*v1)+(fraction of molecules have speed v2*v2)+(fraction of molecules have speed v3*v3)
Similarly we find average speed of the molecules in a gas with the following procedure
We weight each value of v in the distribution,that is we multiply it by the fraction of molecules A(v)v of the molecules
Then we add us all these values of vA(v)dv to get the average speed

So $V_{avg} = \int_{0}^{\infty} vA (v) dv $

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Linked Type Comprehension

The below figure show a hypothetical speed distribution for a sample of N gas particles.

Question 27
What is the relation but a, b & v₀
a. $v_0 (2a+3b) = 2$
b. $v_0 (2a- 3b) = 2$
c. $v_0 (a+ b) = 1$
d. $v_0 (a- b) = 1$

Answer

(a)
Area of curve should be one
so
$ \frac {1}{2}av_0 +av_0 + \frac {1}{2}v_0(b-a)+bv_0=1$
or $v_0 (2a+3b) = 2$

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Question 28
How much fraction of molecules are in speed range v₀ & 3v₀
a. $(\frac {v_0}{2}) (3b+a)$
b. $v_0(3b+c)$
c. $(\frac {v_0}{4}) (3b+a)$
d. $(\frac {v_0}{4}) (3b-a)$

Answer

(a)
Fraction of molecules are in speed range v0 & 3v0 = Area of the curve between the speed range
$=bv_0+av_0+(1/2)v_0(b-a)$
$=(\frac {v_0}{2}) (3b+a)$

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Question 29
What is the average speed of the molecules
a. $\frac {6av_0^2 + 20bv_0^2}{6}$
b.$\frac {6av_0^2 - 38bv_0^2}{6}$
c. $3av_0^2 + 2bv_0^2$
d. $3av_0^2 - 2bv_0^2$

Answer

(a)      V_av = ∫[vA(v)]dv where limits are 0 to infinity
     = ∫[vA(v)]dv where limits are 0 to v₀
+ ∫[vA(v)]dv where limits are v₀ to 2v₀
+ ∫[vA(v)]dv where limits are 2v₀ to 3v₀

     = ∫[v(a/v₀)vdv where limits are 0 to v₀
+ ∫v[{(b-a)/v₀}v+(2a-b)]dv where limits are v₀ to 2v₀
+ ∫[v b]dv where limits are 2v₀ to 3v₀
     = (6av₀² + 20bv₀²)/6

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Question 30
Column A
A. Average energy of monatomic molecule
B. Average energy of Diatomic molecule
C. Average energy of Polyatomic molecule with no vibration
d. Average energy of Diatomic molecule which vibrate

Column B
U. $3K_BT$
V. $\frac {7}{2} K_BT$
X. $\frac {5}{2} K_BT$
Y. $\frac {3}{2} K_BT$
a. A - V,B - X,C - U,D - Y
b. A - Y,B - X,C - U,D - V
c. A -X,B - Y,C - U,D - V
d. A - Y,B - X,C - V,D - U

Answer

(b)

Average energy of monatomic molecule     =(degree of Freedom)*(KT/2)=3KT/2 as degree of Freedom=3
Average energy of Diatomic molecule          =(degree of Freedom)*(KT/2)=5KT/2 as degree of Freedom=5
Average energy of Polyatomic molecule     =(degree of Freedom)*(KT/2)=3KT as degree of Freedom=6
Average energy of Diatomic molecule which vibrate     =(degree of Freedom)*(KT/2)=7KT/2      as degree of Freedom=7

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Question 31
A vessel of volume V contains a mixture of 1 mole of Hydrogen and 1 mole of Oxygen (both considered as ideal). Let $f_1(v)dv$, denote the fraction of molecules with speed between v and (v + dv) with $f_2 (v)dv$, similarly for oxygen. Then
(a) $f_1(v) + f_2(v) =f(v)$ obeys the Maxwell's distribution law.
(b) $f_1 (v)$, $f_2(v)$ will obey the Maxwell's distribution law separately.
(c) Neither $f_1 (v)$, nor $f_2 (v)$ will obey the Maxwell's distribution law.
(d) $f_2 (v)$ and $f_1 (v)$ will be the same.

Answer

(b)

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Question 32
A gas has five molecules having velocities v,2v,3v,4v,5v.Match the column A to column B
Here V_mean is mean velocity and V_rms is root mean square velocity of the molecules

Column A
P. (V_rms)²
Q. (V_mean)²
R (V²)_mean
S. (V²)_mean - (V_rms)²

Column B
W. 9v²
X. 11v²
Y. 10v²
Z. 0

a.P-W,Q-X,R-Y,S-Z
b.P-X,Q-W,R-Z,S-Y
c.P-X,Q-W,R-Y,S-Z
d.P-X,Q-W,R-X,S-Z

Answer

V_mean=(∑ v/N)=[v+(2v)+(3v)+(4v)+(5v)]/5=3v
V_rms=√(∑ v²/N)=√[v²+(2v)²+(3v)²+(4v)²+(5v)²]/5=v(11)^1/2

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Question 33
A gas has $v_0$ as rms at temperature $T_0$ and pressure $p_0$.which of the following is correct
a.If the pressure is doubled keeping temperature constant T0 ,rms will remain same
b.if the mass of the gas molecules is tripled,rms will become .58v0
c. if the temperature is increased such that T=4T0,rms will become 2v0
d. None of the above

Answer

(a),(b) and (c)
V_rms=v₀
V_rms=(3RT/M)^1/2

Now since V_rms depends on temperature and mass
so when pressure is doubled keeping temperature constant, rms remains same
Now .if the mass of the gas molecules is tripled
then
V_rms=(3RT/3M)^1/2=v₀(1/3)^1/2
=.58v₀
Now temperature is increased such that T=4T₀,
V_rms=(3RX4T/M)^1/2=2v₀


V_mean=(∑ v/N)=[v+(2v)+(3v)+(4v)+(5v)]/5=3v
V_rms=√(∑ v²/N)=√[v²+(2v)²+(3v)²+(4v)²+(5v)²]/5=v(11)^1/2

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Question 34
A cubic vessel (with faces horizontal + vertical) contains an ideal gas at NTP. The vessel is being carried by a rocket which is moving at a speed of 200 m/s in vertical direction. The pressure of the gas inside the vessel as observed by us on the ground
(a) remains the same because 200 m/s is very much smaller than $v_{rms}$ of the gas.
(b) remains the same because motion of the vessel as a whole does not affect the relative motion of the gas molecules and the walls.
(c) will increase by a factor equal to $(\frac { v_{rms}^2 + 2500)^2}{v_{rms}^2})$ where $v_{rms}$ was the original mean square velocity of the gas.
(d) will be different on the top wall and bottom wall of the vessel.

Answer

(b)

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Also Read

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• Notes
  • Moleular nature of matter
  • Behaviour Of Gases
  • Kinetic Theory of an ideal gas
  • Pressure of gas
  • Kinetic interpretation of temperature
  • Law of Equipartition of energy
  • Specific Heat Capacity Of Gases
  • Specific heat Capacity of Solids
  • Mean free Path
  • Solved examples
• Assignments
  • Kinetic Theory of Gases Problems
  • Multiple Choice Questions
  • Ideal Gas Laws Problems
• Revision Notes
  • Kinetic Theory of Gases revision sheet

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