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
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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
(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 Solution
$PV = KT$
$PdV = KdT$
$ \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
The work done during the cycle,
a. $PV$
b. $\frac {PV}{2}$
c. $2PV$
d. $\frac {PV}{3}$ Solution
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 Solution
(a)
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. $V = [\frac {2R \Delta T}{ M \sqrt {(\gamma+1)}}]^{1/2}$ Solution
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
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 Solution
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 Solution
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 Solution
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
So
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 Solution
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 Solution
For O2
PV =4RT
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
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. $C_v - (\frac {R^2}{aV})$ Solution
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
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. $4P_0$ Solution
(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. $P_0AL- \frac {3}{2}KL^2$ Solution
at any distance Force on the piston is
F=P0A+3Kx
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$ Solution
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. $R \Delta T + P_0AL$ Solution
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 Solution
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. $1 : n^2 : \gamma ^2$ Solution
We know that
B=-VdP/dV
PVn = constant
differentiating
VndP+nPVn-1dV=0
or -VdP/dV=nP
Similarly for PVγ = Constant
B=γP
And for PV = Constant
B=P
So Ans n:γ:1
Question 19 Find the wok done in the cyclic process as shown in figure
a. $\frac {n^2PV}{2}$
b.$\frac {n^2PV}{8}$
c. $\frac {n^2PV}{16}$
d. none of these Solution
Question 20
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. Solution
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
So
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 Solution
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
ΔU= ΔQ
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 }$ Solution
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
PV=constant
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 Solution
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
So
$\Delta U_A= 1x \frac {5R}{2} (T_f - T_0)$
$\Delta U_B= 1x \frac {3R}{2} (T_f - \frac {7}{3} T_0)$
$ \Delta U_A + \Delta U_B =0$
Solving we get
2Tf-3T0=0
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 Solution
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)}$ Solution
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 = $P_0$
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 = $U_B$ Solution
For Process AB, as temperature remains same, this is isothermal process
So, $U_A = U_B$
Also
$P_0 V_0 = P_B \times 4 V_0$
$P_B = \frac {P_0}{4}$
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 -
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 Solution
(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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