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Physics NEET MCQ
Quiz 6
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Q.1
The change in internal energy, when a gas is cooled from 927°C to 27°C
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a)200%
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b) 100%
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c)300%
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d) 400%
Explanation
We know that Energy of gas E ∝ T Let T1=927 + 273=1200 KT2=27 + 273=300 K Answer:(c)
Q.2
The pressure exerted by an ideal gas at a particular temperature is directly proportional to:
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a) The mean speed of the gas molecules
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b) The mean of the square of speed of the gas molecules
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c) The square of the mean speed of the gas molecules
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d) The root mean square speed of the gas molecule
Explanation
Answer: (b)
Q.3
Figure shows the volume (V) vs Temperature (T) graphs for certain amount of perfect gas at two pressure P1 and P2 Then.
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b) P1 > P2
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c)P1 < P2
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d)P1 ≥ P2
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a)P₁=P₂
Explanation
PV=RT P=RT/V ∴ P ∝ 1/slope More is the slope less is the pressure∴ P1 < P2Answer: (c)
Q.4
A container has an ideal gas at pressure of 5atmosphere and temperature 27°C. One third of the mass of the gas from the container is removed and temperature is increased by 60°C. The pressure of gas in the enclosure is
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a) 5atm
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b) 6 atm
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c)4 atm
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d)3 atm
Explanation
on removal of one third gas number of gas in container is (2/3) of originalFor ideal gas equationAnswer: (c)
Q.5
An ideal gas exerts a pressure P. The mean kinetic energy per unit volume is E. Which of the following relations is correct. [ MLNR 1994]
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a) P=E
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b) P=E/2
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c)P=(2/3)E
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d)P=(3/2)E
Explanation
Average kinetic energy of one mole of gas Answer:(c)
Q.6
the cylinder of capacity 200 litres is filled with hydrogen gas. The total average kinetic energy of translatory motion of molecules is 1.52×105 joules. The pressure of hydrogen in the cylinder is .. [ MPPMT 1993]
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a) 2×105 N/m²
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b) 3×105 N/m²
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c) 4×105 N/m²
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d) 5×105 N/m²
Explanation
200liter=0.2 m3 Pressure P=(2/3) E, here E is energy per unit volume Thus P=(2/3) ( 1.52×105 / 0.2) P=5.06×105 Answer: (d)
Q.7
The molar specific heats of an ideal gas at constantpressure and volume are denoted by Cp and Cv respectively. If γ=Cp/Cv and R is the universal gasconstant, then Cv is equal to … [ NEET 2013]
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a)
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b)
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c)
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d) γR
Explanation
Cp-Cv = R Form given Cp = γCv γCv – Cv = R Cv(γ-1) =R Answer:(b)
Q.8
A piece of iron is heated in a flame. It first becomesdull red then becomes reddish yellow and finallyturns to white hot. The correct explanation for theabove observation is possible by using
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a) Stefan's Law
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b) Wien's displacement Law
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c) Kirchoff's Law
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d) Newton's Law of cooling
Explanation
With change in temperature .light emitted is different or light of different wave length emitted from heated iron road Wien’s displacement law is λmT = constant Answer:(b)
Q.9
A gas is taken through the cycle A → B → C → A,as shown. What is the net work done by the gas?
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a) 2000 J
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b) 1000 J
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c) Zero
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d) –2000 J
Explanation
Work done = area enclosed under the curve Answer:(b)
Q.10
During an adiabatic process, the pressure of a gasis found to be proportional to the cube of its temperature. The ratio of Cp /Cv for the gas is
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a) 4/3
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b) 2
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c) 5/3
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d) 3/2
Explanation
P ∝ T3 PV = nRT PV=C(P)(1/3) ;C=constant P(2/3) V=C PV(3/2)=C On comparing above equation with P (V )γ= Constant We get γ = 3/2 Answer:(d)
Q.11
e given (V – T) diagram, what is the relationbetween pressures P1 and P2?
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b) P2 > P1
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c) P2 < P1
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d) Cannot be predicted
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a) P₂ = P₁
Explanation
PV = nRT Lower the value of P higher the slope. Thus P2 > P1 Answer:(b)
Q.12
The amount of heat energy required to raise the temperature of 1 g of Helium at NTP, from T1K to T2K is [NEET 2013]
0%
a)
0%
b)
0%
c)
0%
d)
Explanation
Amount of energy supplied to 1g of Helium at NTP is utilized for increase in internal energy. Formula for internal energy Degree of freedom f for Helium is 3, number of moles of Helium µ = ¼ Answer:(a)
Q.13
A monoatomic gas at a pressure P, having avolumeVexpands isothermally to a volume 2V and then adiabatically to a volume 16V.The final pressure of the gas is …. (take γ = 5/3)[ AIPMT 2014]
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a) P/ 64
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b) 16P
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c) 64P
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d) 32P
Explanation
Isothermal expansion from V → 2V P1V1 = P2V2 Adiabatic expansion 2V=16V Answer:(a)
Q.14
A thermodynamic system undergoes cyclic process ABCDA as shown in Fig. Theworkdone by the systemin the cycle is …. [ AIPMT 2014]
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a) P0V0 /2
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b) zero
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c) P0V0
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d) 2 P0V0
Explanation
Work in COB = (-P0)V Work in AOD = P0V Total work = Work in COB + Work in AOD =0 Answer:(b)
Q.15
One mole of an ideal diatomic gas undergoes a transition from A to B along a path AB as shown in the figure, The change in internal energy of the gas during the transition is : …[ NEET 205]
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a) -12 kJ
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b) 20 kJ
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c) - 20 kJ
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d) 20 J
Explanation
According to the equipartition theorem the change in internal energy is related to the temperature of the system by for one mole ΔU= CVΔT Now for one mole of gas PaVa = RTa and PbVb = RTb Gas is diatomic CV=5R/2 Answer:(c)
Q.16
A Carnot engine, having an efficiency of η = 0.1 as heat engine, is used as a refrigerator. If the workdone on the system is 10 J, the amount of energy absorbed from the reservoir at lower temperature is :
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a) 1 J
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b) 100 J
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c) 99 J
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d) 90 J
Explanation
Work done = energy absorbed = QH-QL QH = 100 Answer:(b)
Q.17
The ratio of the specific heats γ=CP/CV in terms of degrees of freedom (n) is given by : … [ AIPMT 2015]
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a)
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b)
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c)
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d)
Explanation
Applying first law of thermodynamics to isobaric process we get ΔQ = ΔU + P(V2 – V1) ΔQ = ΔU + PΔV ,But PV = RT for one mole of gas ∴ PΔV= RΔT thus ∴ ΔQ = ΔU + RΔT f is degree of freedom Since dQ/dT is specific heat at constant pressure = CP ∴ CP= CV + R OR CP – CV = R Answer:(d)
Q.18
Two vessels separately contain two ideal gases A and B at the same temperature, the pressure of A being twice that of B. Under such conditions, the density of A is found to be 1.5 times the density of B. The ratio of molecular wigth of A and B is …[ REAIPMT 2015]
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a) 1/2
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b) 2/3
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c) 3/4
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d) 2
Explanation
PV = μRT Number of moles PA= 2PB ;ρA = 1.5ρB Answer:(c)
Q.19
The coefficient of performance of a refrigerator isIf the temperature inside freezer is -20°C, thetemperature of the surroundings to which it rejects heat is : …[ReAIPMT 2015]
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a) 21°C
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b) 31°C
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c) 41°C
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d) 11°C
Explanation
5TH – 5TL = TL 5TH = 6TL TL = 273 -20 =253K 5TH = 6×253 TH = 303.6 K = 30.6°C Answer:(b)
Q.20
An ideal gas is compressed to half its initial volumeby means of several processes. Which of the processresults in the maximum work done on the gas ?
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a) Isothermal
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b) Adiabatic
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c) Isobaric
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d) Isochoric
Explanation
Wadiabatic > WIsothermal > WIsobaric Explanation: 1 ) In isobaric process, volume reduces and pressure remains constant which means PV reduces and hence temperature also reduces. This means that internal energy reduces. This means that part of the work is done by the internal energy. 2) In isothermal process, there is no change in internal energy as the temperature remains constant. This means that the entire work of compression is done by the external agency which is more than the work done in isobaric process. 3) In adiabatic process, there is no heat exchange with the surrounding which means that the temperature and hence internal energy increases. This means that the work is done not only to compression the gas but also to increase the internal energy. Thus, work dne is maximum for adiabatic process than for isothermal process and the least in isobaric process Answer:(b)
Q.21
A refrigerator works between 4째C and 30째C. It is required to remove 600 calories of heat every second in order to keep the temperature of the refrigerated space constant. The power required is: (Take 1 cal = 4.2 Joules)
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a) 2.365 W
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b) 23.65 W
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c) 236.5 W
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d) 2365 W
Explanation
Input in J = 56.36×4.2 = 236.71 J is to be provided in one second Power = 236.5 W Answer:(c)
Q.22
A gas is compressed isothermally to half its initialvolume. The same gas is compressed separatelythrough an adiabatic process until its volume isagain reduced to half. Then :- [ AIPMT 2015]
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a) Compressing the gas isothermally will requiremore work to be done
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b) Compressing the gas through adiabatic process will require more work to be done.
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c) Compressing the gas isothermally oradiabatically will require the same amount ofwork.
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d) Which of the case (whether compressionthrough isothermal or through adiabaticprocess) requires more work will depend uponthe atomicity of the gas.
Explanation
1 ) In isobaric process, volume reduces and pressure remains constant which means PV reduces and hence temperature also reduces. This means that internal energy reduces. This means that part of the work is done by the internal energy. 2) In isothermal process, there is no change in internal energy as the temperature remains constant. This means that the entire work of compression is done by the external agency which is more than the work done in isobaric process. 3) In adiabatic process, there is no heat exchange with the surrounding which means that the temperature and hence internal energy increases. This means that the work is done not only to compression the gas but also to increase the internal energy. Thus, work dne is maximum for adiabatic process than for isothermal process and the least in isobaric process Answer:(b)
Q.23
One mole of an ideal monatomic gas undergoes aprocess described by the equation PV3 = constant.The heat capacity of the gas during this process is…[ NEET II – 2016]
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a) 2 R
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b) R
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c) 3R/2
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d) 5R/2
Explanation
for a process, defined by PVz=constant C is heat capacity, γ is adiabatic exponent Since given gas is monatomic γ = 5/3 and z =3 Answer:(b)
Q.24
The temperature inside a refrigerator is t2°C and the room temperature is t1°C. The amount of heatdelivered to the room for each joule of electricalenergy consumed ideally will be … [ NEET II – 2016]
0%
a)
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b)
0%
c)
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d)
Explanation
Here QL/W is the amount of heat delivered per joule of electric energy= Q Answer:(d)
Q.25
A given sample of an ideal gas occupies a volumeV at a pressure P and absolute temperature T. Themass of each molecule of the gas is m. Which ofthe following gives the density of the gas ?[ NEET II – 2016]
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a) P/(kTV)
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b) mkT
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c) P/(kT)
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d) Pm/(kT)
Explanation
PV= µRT Answer:(d)
Q.26
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.27
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.28
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]
0%
a) P0V0
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b)
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c)
0%
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.29
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.30
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)
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