Multiple choice questions on Kinetic theory of gases for Class 11,JEE and NEET
Multiple choice questions with one or more answer
Question 1 Let A & B are two sample of ideal gases of equal mole .let T be the temperature of both the gas Let $E_A$ and $E_B$ are there total energy respectively .Let $M_A$ and $M_B$ are these respective Molecular Mass .which of these is true
a. $E_A > E_B$
b. $E_A < E_B$
c. $E_A =E_B$
d. none of these
Question 2 An Ideal gas undergoes an state change according to PV diagram. what is the value $V_x$
a. $\frac {V_0}{2}$
b. $V_0$
c. $2V_0$
d. $\frac {V_0}{4}$
As the temperature is constant PV = constant
2P0 Vx = P0V0
Vx = V0/ 2
Linked Comprehensions type
A container contains $N_2$ gas at T K. The no of moles of gas is $n_0$. Consider it behaves like ideal gas. It rms speed is $v_0$
Question 3 What is the total translational kinetic energy
a. $\frac {3}{2} n_0RT$
b. $\frac {3}{2} RT$
c. $\frac {1}{2} RT$
d. $\frac {1}{2} n_0RT$ Solution
KE = (3/2) nRT
= (3/2) n0RT
Question 4 Suppose the temperature of gas is tripled and $N_2$ molecules dissociate into atom. Then what will be the rms speed of atom.
a. $v_0 \sqrt {6}$
b. $ \sqrt {6 v_0}$
c. $v_0 \sqrt {3}$
d. $\sqrt {3 v_0}$ Solution
v0 = √3KT/M
vt = √9KT/(M/2)
=v0√6
Question 5 if $V_p$ denotes most probable velocity of N2 at T then which of these is correct
a. $V_p : V_0= \sqrt {2} : \sqrt {3}$
b. $V_p : V_0= 1 : 1$
c. $V_p : V_0= 2 : 3$
d. $V_p : V_0= \sqrt {2 }: 3$
Question 6 An ideal gas undergoes the process describe by equation
$P = P_0 - aV^2$
Where $P_0$ , a are positive constant and V is the volume of one mole of gas
Maximum temperature attainable by gas
a. $\frac {2}{3} (\frac {P_0}{R}) \sqrt { \frac {P_0}{3a}}$
b. $3 P_0 \sqrt {\frac {P_0}{2Ra}}$
c. $(\frac {1}{3R}) \sqrt {\frac {P_0}{3a}}$
d. $\frac {4}{3} (\frac {P_0}{R}) \sqrt { \frac {P_0}{2}}$
P = P0 - aV2
= P0-a(RT/P)2
PV = RT for one mole of gas
T = (1/R√a )P√(P0 -P)
= (1/R√a )√(P0P2 - P3)
For Tmax
d (P0P2 - P3) /dP= 0
P = 2P0/3
Putting this value
Tmax = 2/3 (P0/R) √(P0/3a)
Question 7 Let V, Vrms and Vp respectively denotes the mean speed, root mean square speed and most probable speed of the molecule in ideal monatomic gas at absolute temperature T. The mass of the molecule
is m then
a. No molecule can have a speed greater the (2)1/2Vrms
b. No molecule can have a speed less the Vp / (2)1/2
c. Vp < V < Vrms
d. The average kinetic charge of a molecule (3/4 )mVp2
So it is clear
Vp < V < Vrms
Also
Average kinetic energy
=(½)mVrms2
Vrms2 = 3KT/M
Vp2 = 2KT/M
Vrms2 = 3/2 Vp2
So E=3/4 Vp2
So Ans. C&D
Linked Comprehensions type
A flask contains Oxygen, Hydrogen & chlorine in the ration of 3:2:1 mixture at 27 °C
Molecular mass of Oxygen = 32
Molecular mass of Hydrogen = 2
Molecular mass of Chlorine = 70.9
Question 8 Find the ratio of average kinetic energy per molecule of Oxygen & Hydrogen
a. 1:2
b. 1:1
c. 1:16
d. 2:1
Question 9 find the ratio of average Kinetic energy per molecule of Hydrogen & chlorine
a. 1:2
b. 1:1
c. 1:16
d. 1:1
All the molecule have same kinetic energy at same temperature
Question 10 find the ratio of mean speed of Oxygen, chlorine, Hydrogen
a. 4:1 : (.45)1/2
b. 5:2 :(5)1/2
c. 1: (.45)1/2:4
d. 2 :(.45)1/2 : (1.5)1/2 Solution
Question 11 Suppose a container is evacuated to have just one molecule of a gas in it. Let Va & Vrms represent the Average Speed and rms speed of the molecule
a. Va > Vrms
b. Va < Vrms
c. Va = Vrms
d. Vrms is undefined
Question 13 Equal Number of molecules of hydrogen & Oxygen are contained in a vessel at one atmosphere
pressure. The ratio of the collision frequency of hydrogen molecules to the of Oxygen molecules on the
container
a. 1:4
b. 4:1
c. 1:16
d. 16:1
n1 = Vrms / 2L ---Hydrogen
n2 = Vrms/2L ---Oxygen
n1/n2 = Vrms of Hydrogen/ Vrms of oxygen
Vrms of Hydrogen/ Vrms of oxygen = (MO/MH)1/2
So n1/n2 = 4:1
Question 14 The prefect gases A, B & C having masses m1, m2& m3 at temperature T1, T2 & T3 are mixed without any loses of internal energy of the molecules. The molecules weight of the gases is M1, M2, M3 what
is the final temperature of mixture.
a. [(m1/M1)T1 + (m2/M2)T2 + (m3/M3)T3]/(m1/M1+m2/M2+m3/M3)
b. [(M1/m1)T1 + (M2/m2)T2 + (M3/m3)T3]/(M1/m1+M2/m2+M3/m3)
c. [(m1)T1 + (m2)T2 +(m3)T3]/(m1+m2+m3)
d. [(M1)T1 + (M2)T2 + (M3)T3]/(M1+M2+M3)
Internal kinetic energy
= (3/2) nRT
so for A KE1 = (3/2 )(m1/M1) RT1,
for B KE2 = (3/2 )(m2/M2) RT2,
for C KE3 = (3/2 )(m3/M3) RT3,
Now final Kinetic energy
KE = 3/2 (m1/M1+m2/M2+m3/M3) RT
Now
KE = KE1 + KE3+KE3
T = [(m1/M1)T1 + (m2/M2)T2 + (m3/M3)T3]/(m1/M1+m2/M2+m3/M3)
Question 15
M moles of a ideal polyatomic gas(Cv=7R/2) are in cylinder at temperature T.A heat Q is supplied to the gas. Some M/3 moles of the gas dissociated into atoms while temperature remains constant. Find the correct relation
(a) $3Q=4MRT$
(b) 2Q=3MRT
(c) $Q=4MRT$
(d) $7Q=4MRT$ Solution
Internal energy before supplying the heat
$= \frac {7}{2} MRT$
Total no of polyatomic moles after the heat is supplied
=2M/3
Total no of monatomic moles after the heat is supplied
=5M/3
Total internal after the split
$= \frac {7}{2} (\frac {2M}{3})RT + \frac {3}{2} (\frac {5M}{3}) RT$
$= \frac {29}{6} MRT$
So
$Q= \frac {29}{6} MRT - \frac {7}{2} MRT$
3Q=4MRT
Ans is a
Question 16
A gas mixture consist of molecules of type A, B, C, D with molecular
masses Ma > Mb > Mc >Md
Two statement are drawn from it Statement A, Average kinetic energy of four type of gases in the mixture are in the ratio
Ea/1 =Eb/1 =Ec/1 =Ed/1 Statement B, Rms speed of molecules of the four types are in the order if V is the rms speed
VD > VC > VC > VA
which one of following is true
a. Only A correct
b. Only B correct
c. A & B both are correct
d. A & B both are wrong
Average kinetic energy is same for all the gases at the same temperature
rms speed is inversely proportional to the Molecular mass of the gas
since Ma > Mb > Mc >Md
VD > VC > VC > VA
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
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
Statement A is Avogadro law. So true
Avogadro no is 6.0255 * 1023 So second one is true
Question 18
There are two statement about Ideal gases Statement A The Vrms of gas molecules depends on the mass of the gas molecule and the temperature Statement B The Vrms 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 Solution
vrms depends on mass and temperature
So A is correct . Ans B is incorrect
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
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
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
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
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 Solution
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
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. Solution
(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.
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.$\sqrt {\frac {8RT}{M}}$ Solution
(a)
This is derived from Maxwell equation.Derivation is beyond the scope of the course
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. $\int_{0}^{\infty}A (v) dv = 0$ Solution
(a)
A(v)dv represent the fraction of molecules...so when all the fractions are considered...this should be equal to 1
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 Solution
(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.
Question 26
What is the formula for Vavg
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 $
(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 $
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. $v_0 (a- b) = 1$ Solution
(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 $v_0 (2a+3b) = 2$
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. $(\frac {v_0}{4}) (3b-a)$ Solution
(a)
Fraction of molecules are in speed range v0 & 3v0 = Area of the curve between the speed range
$=bv_0+av_0+(1/2)v_0(b-a)$
$=(\frac {v_0}{2}) (3b+a)$
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. $3av_0^2 - 2bv_0^2$ Solution
(a)
Vav = ∫[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
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 Solution
(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
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. Solution
(b)
Question 32
A gas has five molecules having velocities v,2v,3v,4v,5v.Match the column A to column B
Here Vmean is mean velocity and Vrms is root mean square velocity of the molecules
Column A
P. (Vrms)2
Q. (Vmean)2
R (V2)mean
S. (V2)mean - (Vrms)2
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 Solution
(a),(b) and (c)
Vrms=v0
Vrms=(3RT/M)1/2
Now since Vrms 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
Vrms=(3RT/3M)1/2=v0(1/3)1/2
=.58v0
Now temperature is increased such that T=4T0,
Vrms=(3RX4T/M)1/2=2v0
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. Solution
(b)
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