Heat and thermodynamics MCQ

In this page we have Multiple Choice questions on heat and thermodynamics for JEE Main and Advanced . Hope you like them and do not forget to like , social share and comment at the end of the page.

Multiple choice questions with one or more answers

Question 1
It is known that curves A, B, C are Isobaric, Isothermal, Adiabatic process then when one is correct
Multiple Choice questions on heat and thermodynamics for Class 11, NEET and JEE
(a) A - Adiabatic, B - Isothermal,, C - Isobaric
(b) A- Isothermal,, B - Adiabatic, C - Isobaric
(c) A - Isobaric, B - Isothermal C - Adiabatic
(d) None of these


Isobaric Pressure is constant
so Curve C is Isobaric
Adiabatic slope is more steep the Isothermal
So Adiabatic - A
And Isothermal - B
so (a) is correct

Question 2
Which of the following graph correctly represent the variation
$\delta = \frac {dV }{V dT}$
for an ideal gas at constant pressure
Multiple Choice questions on heat and thermodynamics for Class 11, NEET and JEE


$PV = KT$
$PdV = KdT$
$ \frac {dV}{dT} = \frac {k}{P}$
$\frac {dV }{ VdT} = \frac {K }{ PV}$
Now as PV=KT
$\frac {dV }{ VdT} = \frac {K }{ KT}= \frac {1}{T}$
or $\delta = \frac {dV }{V dT} = \frac {1}{T}$
So D is Correct

Question 3
An ideal gas taken round the cycle ABCA as shown in PV diagram
Multiple Choice questions on heat and thermodynamics for Class 11, NEET and JEE
The work done during the cycle,
a. $PV$
b. $\frac {PV}{2}$
c. $2PV$
d. $\frac {PV}{3}$


Work done is area of the closed loop
= 1 / 2 x V x P = PV / 2
So b is correct

Question 4
Consider the following statements

(Assertion) The internal energy of an ideal gas does not change during an Isothermal process
(Reason) The decrease in the volume of the gas is compensated by a corresponding increase in pressure when its temperature is constant in accordance with Boyle law

a. Both A & R are true and R is correct Explanation of A
b. Both A & R are true and R is not correct Explanation of A
c. A is true R is false
d. A is false but R is true



Question 5
A thermally insulated vessel containing an gas when molar mass is M and Ratio of specific heat Cp / Cv = γ move with a velocity V.The gas temperature increase by ΔT due to sudden stoppage of vessel. Find the value of V in terms of ΔT, M, γ
a. $V = [\frac {2R \Delta T} {M (\gamma-1)}]^{1/2}$
b. $V = [\frac {2R \Delta T}{ M (\gamma + 1)}]^{1/2}$
c. $V = [\frac {2R \Delta T}{ M \sqrt {(\gamma-1)}}]^{1/2}$
d. $V = [\frac {2R \Delta T}{ M \sqrt {(\gamma+1)}}]^{1/2}$


From energy Conservation
$ \Delta U = \frac {1}{2} nMV^2$ where n = moles of gas
now $\Delta U = \frac {nR \Delta T} {\gamma-1}$
So $\frac {1}{2} nMV^2=\frac {nR \Delta T} {\gamma-1}$
$V^2 = \frac {2R \Delta T }{ M (\gamma-1)}$
$V = [\frac {2R \Delta T} {M (\gamma-1)}]^{1/2}$

Question 6
Match the column
Multiple Choice questions on heat and thermodynamics for Class 11, NEET and JEE
Multiple Choice questions on heat and thermodynamics for Class 11, NEET and JEE
a. a-> q, b -> p, c-> r
b. a-> p, b -> q, c-> r
c. a-> r, b -> q, c-> p
d. a-> p, b -> r, c-> q


a - AB
b - AC
c - BC

Question 7
An ideal gas is taken through a cyclic thermodynamics process through four steps.
The amount of heat involved in the steps are Q1 = 5960 J,Q2 = - 5600 J,Q3= -3000 J,Q4 = -3600 J
respectively. The corresponding quantities of Internal energy changes are ΔU1 = 3.760 J ,ΔU2 = - 4800 J,ΔU3 = -1800 J,ΔU4 = ?
find the value ΔU4 & net work done
a. 2930 J, 960 J
b. 2830 J, 900 J
C. 2930 J, -960 J
d. -2930 J, 960 J


since in cycle process total internal change is zero.
$\Delta U_1 + \Delta U_2 + \Delta U_3 + \Delta U_4 = 0$
$3670 - 4800 - 1800 + \Delta U_4 = 0$
$ \Delta U_4 = 2930$ J
Now in Cycle
$ \Delta W = \Delta Q$
= 5960 - 5600 - 3000 + 3600 = 960 J.
Hence answer is a

Question 8
An ideal gas who ratio of specific heat Cp / Cv = γ is Expanded according to the law P = av2
when a is constant. The ratio of final volume to Initial value is n .find the ΔU.Initial volume is V0
a. aV03 (n3-1) / γ-1
b. aV02 (n2 - 1) / γ-1
c. aV03 (n3 + 1) /γ+1
d. aV02 (n2 + 1) /γ-1


ΔU = nCvΔT
= (nR / γ-1)(T2-T1)
= P2V2 - P1V1 / γ-1
Now P1 = aV12
P2 = aV22
Now V1=V0
V2/ V1 = n
So V2=nV0
P1 = aV02
P2 = an2V02
Substituting these values
ΔU =aV03 (n3-1) / γ-1

Question 9
During an adiabatic process the square of the pressure of a gas is proportional to the fifth power of its absolute temperature. The ratio of specific heat Cp / Cv for that gas is
a. 3/5
b. 4/3
c. 5/3
d. 3/2


P2= k T5
P2 =k(PV/nR)5
P2 = k'P5V5 where k'=k/(nR)5 which is a constant
P3V5 = constant
PV5/3 = constant
so Cp / Cv = 5/3

Question 10
A vessel contains 4 mole of O2(relative molar mass 32) at temperature T. The pressure is P. An identical vessel contains 1 mole of nitrogen of temperature 2T. find the pressure the
a. P/2
b. 2P
c. 8P
d. P


For O2
V = 4RT / P
For nitrogen
P'V = 2RT
P' = 2RT/V=P/2
a is correct

Question 11
What is the molar specific heat of a isothermal & adiabatic process respectively
a. ∞ , 0
b. 0, ∞
c. 0, 0
d. none of these


a, ∞ , 0

Question 12
An ideal gas has molar specific heat at constant volume = Cv. find the molar heat capacity of this gas as a function of volume V at the gas goes through process.
$T = T_0(e^{aV} + 1)$
a. $C_v + (\frac {R}{aV}) + (\frac {R}{aVe^{aV}})$
b. $C_v + (\frac {R}{aV})$
c. $C_v + (\frac {R^2}{aV})$
d. $C_v - (\frac {R^2}{aV})$


Molar specific heat as per the formula so
C = dQ/ ndT
= (nCvdT+PdV)/ndT
= Cv + (RT/V)(dV/dT)
as PV=nRT so P/n=RT/V Now , T = T0(eaV + 1)
dT = T0aeaVdV
dV/dT = 1/T0aeaV
So C = Cv + (RT/V)( 1 / T0aeaV)
= Cv + (R/V)T0(1 + eaV)/T0aeaV
= Cv + (R/aV) + (R/aVeaV)

Linked Comprehension type

Multiple Choice questions on heat and thermodynamics for Class 11, NEET and JEE
1 mole of a diatomic idea gas is enclosed in a adiabatic cylinder filled with a smooth light adiabatic Piston. The Piston is connected to three spring of spring constant K as shown in figure. The area of cross-section of Cylinder is A Initially spring is in its natural length and atmosphere pressure is P0

Question 13
find the pressure of the gas
a. $P_0$
b. $2P_0$
c. $\frac {P_0}{2}$
d. $4P_0$


(a) as Gas pressure is equal to atmosphere pressure

Question 14
If the heat is supplied to the gas and piston move by distance L due to that then what is the work done by the gas
a. $P_0AL+ \frac {3}{2}KL^2$
b. $P_0AL$
c. $\frac {3}{2}KL^2$
d. $P_0AL- \frac {3}{2}KL^2$


at any distance Force on the piston is
so Work done =∫Fdx=∫ (P0A+3Kx)dx
Integrating from 0 to L
=P0AL+ (3/2)KL2


Question 15
if Temp changes by ΔT due to heat transfer and γ=5/2 then Find out the internal energy change
a. $R \Delta T$
b. $ \frac {2}{3}R \Delta T$
c.$ \frac {5}{2} R \Delta T$
d. $ \frac {1}{2} R \Delta T$


ΔU = nCvΔT
= [1xR/(γ-1 )]ΔT
= (2/3)RΔT


Question 16
Find the total heat supply
a. $R \Delta T +\frac {3}{2}KL^2$
b. $P_0 AL+\frac {3}{2}KL^2 + \frac {2}{3} R \Delta T$
c. $R \Delta T -\frac {3}{2}KL^2$
d. $R \Delta T + P_0AL$


Q= ΔU+ W
= P0AL+ (3/2)KL2+ (2/3)RΔT


Multiple choice questions with one or more answers

Question 17
One mole of ideal gas having adiabatic coefficient γ1 is mixed with 1 one mole of an ideal gas have adiabatic coefficient γ2
Find the γ of the mixture
a.$ \frac {(2 \gamma _1 \gamma _2 - \gamma _1 - \gamma _2 ) }{( \gamma _1 + \gamma _2 - 2)}$
b.$ \frac {(2 \gamma _1 \gamma _2 - \gamma _1 - \gamma _2 ) }{( \gamma _1 + \gamma _2 + 2)}$
c.$ \frac {(2 \gamma _1 \gamma _2 + \gamma _1 - \gamma _2 ) }{( \gamma _1 + \gamma _2 - 2)}$
d. None of these


(1+1) CvdT = [1 x R /(γ1-1)] dT + [1x R / (γ2-1)dT]
Cv = (R / 2) [(γ1 + γ2 - 2)/(γ1-1)(γ2-1)]
Cp= Cv + R
= R [(γ1 + γ2 - 2 + 2γ1 γ2 -2γ1 - 2γ2 + 2) / 2 (γ1 -1) (γ2 -1)]
= R (2γ1γ11 - γ2 ) / 2 ((γ1-1) (γ2-1)
γ= (2γ1γ11 - γ2 ) /(γ1 + γ2 - 2)


Question 18
We have a process defined as $PV^n = constant$
and we have an adiabatic process defined by $PV^{\gamma} = Constant$
and so thermal process defined as $PV = Constant$
find the Ratio of Bulk modules of Poly-tropic, adiabatic, isothermal process
a. $n: \gamma : 1$
b. $1 : n : \gamma$
c. $n^2 : \gamma ^2 : 1$
d. $1 : n^2 : \gamma ^2$


We know that
PVn = constant
or -VdP/dV=nP

Similarly for PVγ = Constant
And for PV = Constant

So Ans n:γ:1



a. $\frac {n^2PV}{2}$
b.$\frac {n^2PV}{8}$
c. $\frac {n^2PV}{16}$
d. none of these

Question 20
Multiple Choice questions on heat and thermodynamics for Class 11, NEET and JEE
Let Q1, Q2 , Q2 heat given to the system in processes A, B, C respectively
consider two statements

a, Q1 > Q2 > Q3
b, WA > WB > WC

1, Both A & B are Correct
2, Both A & B are wrong
3. A is Correct Only
4. B is Correct Only.


Now we know that
ΔU = Q - W
for three process
ΔU = Q1 - WA Q1 = ΔU + WA
ΔU = Q2 - WB Q2 = ΔU +WB
ΔU = Q3 - WC Q3 = ΔU + WC
Now Work done is given by Area under curve so WA > WB > WC

Q1 > Q2 > Q3
So A & B both are correct


Question 21
Match the Column
Column A
a. Isothermal process
b. Adiabatic Process
c. Isobaric process
d. Isochoric process
Column B
x. ΔU = ΔQ
y. ΔU = ΔQ - ΔW
z. ΔU = -ΔW
w. ΔQ = ΔW

a, a -> x, b-> y, c->z, d->w
b, a->w, b->z, c->x, d->y
c, a->y, b->x, c->z, d->w
d, a->z,b->w,c->y,d->x


 Isothermal Process
ΔU = 0
so ΔQ =ΔW
Isochoric Process
ΔU = ΔQ - ΔW
Adiabatic Process

ΔQ = O
so ΔU = -ΔW
Isobaric process
V = Const
so ΔW = 0

Question 22
The ratio of adiabatic bulk modulus and isothermal bulk modulus of a gas ($\gamma = \frac {C_p}{C_v}$) is
a. $ \frac {\gamma -1}{\gamma}$
b. 1

c. $\gamma $
d. $ \frac {\gamma}{\gamma -1 }$


Bulk Modulus is defined as
$B=-\frac {V \Delta P}{\Delta V}$
For adiabatic process
$PV^{\gamma }=Constant$
Differentiating partially we get
$\Delta P V^{\gamma } + P \gamma V^{\gamma  -1} =0$
or  $- \frac {V \Delta P}{\Delta V} =\gamma P$
For isothermal Process
or $  - \frac {V \Delta P}{\Delta V}=P$
So ratio is equal to $\gamma $

Question 23
Two boxes A and B containing different ideal gases are placed on table
Box A contain one mole of gas m where (Cv=5R/2) at Temperature T0
Box B contains one mole of gas n where (Cv=3R/2) at temperature (7/3) T0
The boxes are then put into thermal contact with each other and heat flows between until the gases reach a common final temperature Tf
Which of the following relation is correct?
a. 2Tf-3T0=0
b. 2Tf-7T0=0
c. 2Tf-5T0=0
d. Tf-3T0=0


Change in the internal energy of the system is zero i.e increase in internal of one gas is equal to      decrease in internal energy of other
$\Delta U_A= 1x \frac {5R}{2} (T_f - T_0)$
$\Delta U_B= 1x \frac {3R}{2} (T_f - \frac {7}{3} T_0)$
$ \Delta U_A + \Delta U_B =0$
Solving we get
So a is correct

Question 24
Which one of the following statement is true about a gas undergoing isothermal change
a. The temperature of the gas is constant
b. The pressure of the gas  remains constant
c. the volume of the gas remains constant
d. The gas is completely insulated from the surrounding’s


Correct ans is a

Question 25
Three copper blocks of masses $M_1$, $M_2$ and $M_3$ kg respectively are brought into thermal contact till they reach equilibrium. Before contact, they were at $T_1$, $T_2$, $T_3$ ($T_1 > T_2 > T_3$ ). Assuming there is no heat loss to the surroundings, the equilibrium temperature T is (s is specific heat of copper)
a. $ \frac {T_1 + T_2 + T_3}{3}$
b. $ \frac {M_1T_1 + M_2T_2 + M_3T_3}{M_1 + M_2 + M+3}$
c. $ \frac {M_1T_1 + M_2T_2 + M_3T_3}{3(M_1 + M_2 + M+3)}$
d. $ \frac {M_1T_1s + M_2T_2s + M_3T_3s}{3(M_1 + M_2 + M+3)}$


Let the equilibrium temperature of the system = T
Let $T_1, T_2 < T < T_3$
As there is no loss to the surroundings.
heat lost by $M_3$ = Heat gain by $M_1$ + Heat gain by $M_2$
$M_3s(T_3 T) =M_1s(T T_1) + M_2s(T T_2)$
$M_3sT_3 M_3sT = M_1sT M_1sT_1 + M_2sT M_2sT_2$
$T(M_3 + M_1 + M_2) = [M_3T_3 + M_1T_1 + M_2T_2]$
$ T=\frac {M_1T_1 + M_2T_2 + M_3T_3}{M_1 + M_2 + M+3}$
Hence verifies option (b).

Question 26
One mole of an ideal gas goes through the cyclic process ABCA. Pressure at State A = $P_0$
Multiple Choice questions on heat and thermodynamics for Class 11, NEET and JEE
Which of the following is correct
a. Pressure at C is $\frac {P_0}{4}$
b. temperature at C is $\frac {T_0}{4}$
c. $W_{AB}=P_0V_0 ln 4$
d. $U_A = $U_B$


For Process AB, as temperature remains same, this is isothermal process
So, $U_A = U_B$
$P_0 V_0 = P_B \times 4 V_0$
$P_B = \frac {P_0}{4}$
Also $W_{AB} = n RT_0 ln {V_2}{V_1} = P_0V_0 ln 4$
The process BC is not clear as it is not mentioned line BC passes through origin, so nothing can be said about state C
So (c) and (d) are correct

Question 27
The figure shows the P-V plot of an ideal gas taken through a cycle ABCDA. The part ABC is a semi-circle and CDA is half of an ellipse. Then -
Multiple Choice questions on heat and thermodynamics for Class 11, NEET and JEE
a. the process during the path A ? B is isothermal
b. heat flows out of the gas during the path B ? C ? D
c. work done during the path A ? B ? C is zero
d. positive work is done by the gas in the cycle ABCDA


(B) $ \Delta Q = \Delta U + W$
In BCD : W is negative
$ \Delta U =\frac {P_2V_2 - P_1V_1}{\gamma -1 }$
= -ve
(D) Cycle is clockwise.

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