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Physics NEET MCQ
Heat And Thermodynamics Mcq
Quiz 7
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Q.1
Thermodynamic processes are indicated in the following diagram. Match the following
Column-1
Column-2
P. Process I
a. Adiabatic
Q. Process II
b. Isobaric
R. Process III
c. Isochoric
S. Process IV
d. Isothermal
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a) a) P → a, Q → c, R → d, S → b
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b) P → c, Q → a, R → d, S → b
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c) P → c, Q → d, R → b, S → a
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d) P → d, Q → b, R → a, S → c
Explanation
Process I = Isochoric Process II = Adiabatic Process III = Isothermal Process IV = Isobaric Answer:(b)
Q.2
A Carnot engine having an efficiency of 1/10 as heat engine, is used as a refrigerator. If the work done onthe system is 10 J, the amount of energy absorbed from the reservoir at lower temperature is … [ NEET 2017]
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a) 1 J
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b) 90 J
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c) 99 J
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d) 100 J
Explanation
TH - TL=W 100-TL =10 TL=90 Answer:(b)
Q.3
The given below p-v diagram represents the thermodynamic cycleof an engine, operating with an ideal monoatomic gas. Theamount of heat, extracted from the source in a single cycle is … [ IIT Mains 2013]
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a) P0V0
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b)
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c)
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d) 4P0V0
Explanation
Q = U-W For Path AB , W = 0 Q = U ΔU= µCVΔT But PV =µRT Therefore ΔP(V) = µRΔT on substitution we get For B to C Work = 2P0V0 As f = 3 But PV =µRT Therefore P(ΔV) = µRΔT on substitution we get Thus total heat extracted = Answer:(b)
Q.4
Two non-reactive monoatomic ideal gases have their atomic masses in the ratio 2 :The ratio of their partial pressures, when enclosed in a vessel kept at a constant temperature, is 4 :The ratio of their densities is
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a) 1 : 4
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b) 1 : 2
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c) 6 : 9
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d) 8 : 9
Explanation
Both the cases are at same temperature and are in same vessel. Thus Answer:(d)
Q.5
Paragraph One mole of a monatomic ideal gas is taken along two cyclic processes E → F → G → E and E → F → H → E as shown in the PV diagram. The processes involved are purely isochoric, isobaric, isothermal or adiabatic. Match the paths in List I with the magnitudes of the work done in List II and select the correct answer using the codes given below the lists.
List I
List II
P. G → E
160 P
0
V
0
1n2
Q. G → H
36 P
0
V
0
R. F → H
24 P
0
V
0
S. F → G
31 P
0
V
0
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a) P → 4, Q → 3, R → 2, S → 1
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b) P → 4 , Q → 3, R → 1, S → 2
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c) P → 3, Q → 1, R → 2 S → 4
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d) P → 1, Q → 3 , R → 2 , S → 4
Explanation
Standard graph for process From the graph F→ G is isothermal process 32P0 V0=PO VG VG = 32P0 From the graph F→ H is adiabatic process As γ=5/3 8V0 = VH P) G→ Eiso baric process WGE = P0 (VE − VG ) = P0(V0 − 32V0 ) = −31P0V0 Thus P → 4 Q) G→ Hiso baric process WGH=P0 (8V0 − 32V0 ) = −24P0V0 Thus relate to Q → 3 R) F→ H adiabatic process Thus R → 2 S) F→ G is isothermal process WFG=160P0 V0 Thus S → 1 P → 4,Q→ 3, R → 2, S → 1 which is in option a Answer:(a)
Q.6
PARAGRAPH In the figure a container is shown to have a movable (without friction)piston on top. The container and the piston are all made of perfectlyinsulating material allowing no heat transfer between outside and inside thecontainer. The container is divided into two compartments by a rigidpartition made of a thermally conducting material that allows slow transferof heat. The lower compartment of the container is filled with 2 moles of an ideal monatomic gas at 700 K and the upper compartment is filled with 2moles of an ideal diatomic gas at 400 K. The heat capacities per mole of an ideal monatomic gas are Cv=3R/2 , Cp= 5R/2 and thosefor an ideal diatomic gas are CvM = 5R/2 , Cp=7R/2 . 187A) Consider the partition to be rigidly fixed so that it does not move. When equilibrium isachieved, the final temperature of the gases will be
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a) 550 K
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b) 525 K
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c) 513 K
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d) 490 K
Explanation
Now partition can move so during transfer of heat pressure will remain constant So we will use value of CP for both n1 Cp (700-T) = n2 Cp (T-400) 3500-5T=7T-2800 12T= 6300 T = 525K Work PΔV =nRΔT Work done by upper diatomic gas w1 = n1R ΔT1 = 2(R) 125 = 250 R Work done by lower monoatomic gas w2 = n2 R ΔT2 = 2 (R) (−175) = − 350 R Net work done = -350R +250R= -100R Answer:(d)
Q.7
than one correct answer Q188) A container of fixed volume has a mixture of one mole of hydrogen and one mole of heliumin equilibrium at temperature T. Assuming the gases are ideal, the correct statement(s)is(are)
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a) The average energy per mole of the gas mixture is 2RT
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b) The ratio of speed of sound in the gas mixture to that in helium gas is √(6/5)
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c) The ratio of the rms speed of helium atoms to that of hydrogen molecules is 1/2
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d) The ratio of the rms speed of helium atoms to that of hydrogen molecules is 1/√2
Explanation
Option a Thus average energy per mole = 2RT option “a” correct Option b ratio of speed of sound in the gas mixture option "d" Velocity of sound in gas For mixture of gases Cp = Cpm + Cpd Cv = Cvm + Cvd Helium Option b is correct Answer:(a, b, d)
Q.8
An ideal monoatomic gas is confined in a horizontal cylinder by a spring loaded piston (as shown in the figure). Initially the gas is at temperature T1, pressure P1 and volume V1 and the spring is in its relaxed state. The gas is then heated very slowly to temperature T2, pressure P2 and volume VDuring this process the piston moves out by a distance x. Ignoring the friction between the piston and the cylinder, the correct statement(s) is(are) ... [IIT Advance 2015]
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a) If V2 = 2V1 and T2 = 3T1, then the energy stored in the spring is P1 V1/4
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b) If V2 = 2V1 and T2 = 3T1, then the change in internal energy is 3P1V1
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c) If V2 = 3V1 and T2 = 4T1, then the work done by the gas is 7P1V1/3
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d) If V2 = 3V1 and T2 = 4T1, then the heat supplied to the gas is 17 P1V1/6
Explanation
Work done by gas = Potential energy of spring Let surface area of piston be S, then change in volume of gas ΔV = Sx Force due to spring = kx Thus pressure on piston when piston displaced by x, P2= kx/S = k x2/ΔV ∴ P2ΔV=kx2 Potential energy of spring = (1/2) kx2 Work done by gas = P2 ΔV/2 Option a ΔV= V2 - V1 = 2V1 - V1 = V1 Work done by gas option A is wrong Option B Change in internal energy As f =3 for mono atomic gas ( degree of freedom) and by ideal gas equation As calculated P2=3P1/2 Option “b” is correct Option “c” Work done by gas = P.E. of spring . Thus option “c” wrong Option d Heat supplied = Work done by gas + change in internal energy option “d” wrong Answer:(b)
Q.9
One mole of an ideal gas at 300 K in thermal contact with surroundings expands isothermally from 1.0 L to 2.0 L against a constant pressure of 3.0 atm. In this process, the change in entropy of surroundings (ΔSsurr) in JK-1 is (1 L atm = 101.3 J) [IIT Advance 2016]
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a) 5.763
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b) 1.013
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c) -1.013
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d) -5.763
Explanation
ΔE = q + w 0 = q - Pext ΔV q = Pext ×V = 3 atm (2 - 1) L = 3 atm L q= (3 × 101.3) Joule Answer:(c)
Q.10
A gas is enclosed in a cylinder with a movable frictionless piston. Its initial thermodynamic state at pressure Pi = 105 Pa and volume Vi = 10-3 m3 changes to a final state at Pf = (1/32) × 105 Pa and Vf = 8 × 10-3 m3 in an adiabatic quasi-static process, such that P3V5 = constant. Consider another thermodynamic process that brings the system from the same initial state to the same final state in two steps: an isobaric expansion at Pi followed by an isochoric (isovolumetric) process at volume Vf. The amount of heat supplied to the system in the two-step process is approximately … [ IIT Advance 2016]
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a) 112 J
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b) 294 J
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c) 588 J
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d) 813 J
Explanation
For the Adiabatic process A to C P2V5 = constant thus γ = 5/2, monoatomic gas For process A to B pressure is constant Heat Q1 Q1 = nCPΔT Q1 = 1750 J For process B to C , volume is constant Heat Q2 Q2 = nCvΔT Q2 = -1162.5 J net Q= 1750 J -1162.5 J = 588 J Answer:(c)
Q.11
PARAGRAPH Answer 192A, 192B and 192C by appropriately matching the information given in the three columns of the following table. An ideal gas is undergoing a cyclic thermodynamics process in different ways as shown in the corresponding P–V diagrams in column 3 of the table. Consider only the path from state 1 to stateW denotes the corresponding work done on the system. The equations and plots in the table have standard notations as used in thermodynamics processes. Here γ is the ratio of heat capacities at constant pressure and constant volume. The number of moles in the gas is n. Q192A) Which of the following options is the only correct representation of a process in which ΔU = ΔQ – PΔV?
Column-1
Column-2
Column-3
(i) Isothermal
(ii) Isochoric
(iii) Isobaric
(iv) Adiabatic
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a) (II) (iv) (R)
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b) (II) (iii) (P)
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c) (II) (iii) (S)
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d) (III) (iii) (P)
Explanation
Now partition can move so during transfer of heat pressure will remain constant So we will use value of CP for both n1 Cp (700-T) = n2 Cp (T-400) 3500-5T=7T-2800 12T= 6300 T = 525K Work PΔV =nRΔT Work done by upper diatomic gas w1 = n1R ΔT1 = 2(R) 125 = 250 R Work done by lower monoatomic gas w2 = n2 R ΔT2 = 2 (R) (−175) = − 350 R Net work done = -150R Answer:(a)
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