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Heat And Thermodynamics Mcq
Quiz 2
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Q.1
An ideal gas heat engine operates in a Carnot cycle between 227°C and 127°C. It absorbs 6kcal at the higher temperature. The amount of heat (in kcal) converted into work is equal to .. [ CBSE-PMT 2003]
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a) 1.2
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b) 4.8
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c)3.5
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d)1.6
Explanation
Efficiency η=1 -( T2 - T1)T1=227+273=500KT2=127+273=400Kby substituting values we get η=1/5Hence output work=η × HEat input=6/5=1.2kcalAnswer: (a)
Q.2
In thermodynamic process which of the following statement is not true [ CBSE-PMT 2009]
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a) In an isochoric process pressure remains constant
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b) In an isothermal process the temperature remains constant
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c) In an adiabatic process PVγ=constant
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d)In an adiabatic process the system is insulated from the surroundings
Explanation
In an isochoric process volume remains constant whereas pressure remains constant in isobaric process. Answer:(a)
Q.3
An deal gas undergoing adiabatic change has the following pressure-temperature relationship [ CBSE-PMT 1996]
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a)
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b)
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c)
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d)
Explanation
We know that in adiabatic process PVγ=constant From ideal gas equation, we know that PV=nRT V=nRT/P Putting the value of V in above equation we get P(1-γ) Tγ=constantAnswer: (d)
Q.4
A gas at 27°C temperature and 30 atmospheric pressure is allowed to expand to the atmospheric pressure. If the volume becomes 10 times its initial volume, then the final temperature becomes [ CBSE-PMT 2001]
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a)100°C
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b) 173°C
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c)273°C
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d)-173°C
Explanation
Initial temperature T1=27°C=300KInitial pressure P1=30 atmInitial Volume V1=V Final Pressure P2=1atm Final volume V2=10VWe know form the general gas equation thatBy substituting the values in above equation and on solving we get T2=100K=-173°CAnswer: (d)
Q.5
One mole of an ideal gas at an initial temperature of T K does 6R joules of work adiabatically. If the ratio of specific heats of this gas at constant pressure and at constant volume is 5/3, the final temperature of gas will be... [ CBSE-PMT 2004]
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a) (T-4) K
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b) (T+2.4)K
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c)(T-2.4)K
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d)(T+4)K
Explanation
Work for Adiabatic process is given byNow P1V1=nRT1P2V2=nRT2Thus n=1, T1 W=6R and γ=5/3 in above equation we getAnswer: (a)
Q.6
During an isothermal expansion, a confined ideal gas does -150 J of work against its surroundings. This implies that.. [ CBSE-PMT 2011]
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a) 150 J heat has been removed from the gas
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b) 300 J of heat has been added to the gas
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c)no heat is transferred because the process is isothermal
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d)150J of heat has been added to the gas
Explanation
in question work done by gas is given -150J, so that according to it, answer will be (d) Answer:(d)
Q.7
The molar specific heat at constant pressure of an ideal gas is (7/2)R. The ratio of specific heat at constant pressure to that at constant volume is .. [ CBSE-PMT 2006]
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a) 8/7
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b) 8/7
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c) 9/7
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d) 7/5
Explanation
Cp=(7/2)R Cv=Cp - R ∴ Cv=(7/2)R-R=(5/2)R Cp / Cv=7/5Answer: (d)
Q.8
First law of thermodynamics is consequence of conservation of [ CBSE-PMT 1988]
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a)work
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b) energy
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c)heat
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d)all of these
Explanation
Answer: (b)
Q.9
In for a gas R/Cv=0.67, the gas is made up of molecules which are [ CBSE-PMT 1992]
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a) diatomic
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b) mixture of diatomic and polyatomic molecules
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c)monoatomic
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d)polyatomic
Explanation
since R / Cv=0.67 ∴ Cv=(3/2)R, hence gas is monoatomic Answer: (c)
Q.10
If γ be the ratio of specific heats of a perfect gas, the number of degrees of freedom of a molecule of gas is .. [ CBSE-PMT 2000]
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a)
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b)
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c)
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d)
Explanation
Ratio of specific heat=1 + (2/n), here n is degree of freedomγ = 1+ (2/n) ∴ n=2 / (γ - 1) Answer:(c)
Q.11
An ideal gas at 27°C is compressed adiabatically to 8/27 of its original volume. The rise in temperature is ( γ=5/3) [ CBSE-PMT 1999]
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a) 475°C
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b) 402°C
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c) 275°C
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d) 175°C
Explanation
T=27°C=300K γ=5/3; V2=(8/27) V1 ; thus V1 / V2=27/8 From adiabatic process we know that T1V1γ-1=T2V2γ-1 T2=675-273=402°C Answer: (b)
Q.12
The first law of thermodynamics expresses [ MPPMT 1987]
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a)law of conservation of momentum
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b) law of conservation of energy
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c)law of conservation of mass
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d)all
Explanation
Answer: (b)
Q.13
Which of the following formula is wrong?
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a)
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b)
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c)
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d)
Explanation
Answer: (d)
Q.14
A gas at N.P.T is suddenly compressed to one-fourth of its original volume. If γ is supposed to be (3/2), then the final pressure is [ MPPMT 1991[
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a) 4 atmospheres
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b) (3/2) atmospheres
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c)8 atmospheres
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d)(1/4) atmospheres
Explanation
For adiabatic expansion Answer:(c)
Q.15
The first law of thermodynamics is the special case of ..[ CPMT 1985]
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a) Newton's law
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b) the law of conservation of energy
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c) Charlie's law
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d) the law of heat exchange
Explanation
Answer: (b)
Q.16
For an adiabatic expansion of perfect gas the value of (ΔP/P) is equal to ..[ PMT 1990]
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a)-γ1/2(ΔV / V)
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b) (ΔV / V)
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c)-γ(ΔV / V)
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d)-γ2(ΔV / V)
Explanation
For adiabatic expansionAnswer: (c)
Q.17
A gas has .. [ CPMT 1988]
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a) one specific heat only
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b) two specific heats only
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c)infinite number of specific heats
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d)no specific heat
Explanation
Answer: (c)
Q.18
A Carnot's engine takes 300 calories of heat at 500K and rejects 150 calories of heat to sink. The temperature of the sink is.. [ MPPMT 1990]
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a) 1000 K
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b) 750 K
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c)250 K
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d)125K
Explanation
We know that efficiency of heat engines e=1 -(Q2 / Q1) --eq(1) Here Q1 is heat absorbedQ2 is heat rejected Alsoe=1 - (T1 / T2) --eq(2)Here T1 is temperature of sinkT2 is temperature of source from equation 1 and 2 (Q2 / Q1)=(T1 / T2) 150 / 300=T1 / 500T1=250 K Answer:(c)
Q.19
The source and sink temperature of a carnot engine are 400K and 300K, respectively. What is its efficiency? [ NCRT 1990]
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a) 100%
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b) 75 %
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c) 33.3 %
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d) 25 %
Explanation
e=1 - (T1 / T2) Here T1 is temperature of sinkT2 is temperature of source e=1 - 300 /400 e=0.25 % efficiency=25% Answer: (d)
Q.20
The slopes of isothermal and adiabatic curves are related as ..
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a)isothermal curve slope=adiabatic curve slope
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b) isothermal curve slope=γ × adiabatic curve slope
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c)γ × isothermal curve slope=adiabatic curve slope
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d)(1/2) × isothermal curve slope=adiabatic curve slope
Explanation
For isothermal curve slope For adiabatic curve slope From above adiabatic curve slope=γ × isothermal curve slopeAnswer: (c)
Q.21
If γ denotes the ratio of the two specific heats of gas, the ratio of the slopes of adiabatic and isothermal P-V curves at their point of interaction is : [ NCRT 1990]
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a) 1 / γ
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b) γ
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c)γ - 1
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d)γ + 1
Explanation
Answer: (b)
Q.22
In reversible isochoric change [ NCRT 1990]
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a) ΔW=0
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b) Δ Q=0
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c)ΔT=0
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d)ΔU=0
Explanation
For isochoric process volume don't change Answer:(a)
Q.23
If R=universal gas constant , the amount of of heat needed to rise the temperature of 2 moles of ideal monoatomic gas from 273 K to 37K when no work is done [ MPET 1990]
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a) 100R
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b) 150R
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c) 300R
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d) 500R
Explanation
When no work is done gas is monoatomic thus Cv=(3/2)R dQ=dU=µ CvdT dQ=2 × (3/2)R ×( 373 -273)=300R Answer: (c)
Q.24
A sample gas expands from volume V1 to VThe amount of work done by the gas is greatest when the expansion is ..[ CBSE 1997]
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a)isothermal
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b) isobaric
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c)adiabatic
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d)equal in all case
Explanation
When pressure is constant ΔU=0 and ΔQ=WAnswer: (b)
Q.25
Find the change in internal energy of the system when a system absorbs 2 kilocalorie's of heat and at the same times does 500 joules of work [ Andhra CET]
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a) 7900 J
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b) 8200 J
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c)5600 J
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d)6400 J
Explanation
2kilocalories=2000 × 4.2=8200 joules From first law of thermodynamics ΔU=ΔQ - W ΔU=8200 - 500=7900 joulesAnswer: (a)
Q.26
The temperature of 5 moles of gas which was held at constant volume was changed fro 100°C to 120°C. The change in the internal energy of the gas was found to be 80 joule, the total heat capacity of the gas at constant volume will be equal to [ CPMT 1988]
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a) 8 joule per K
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b) 0.8 joule per K
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c) 4.0 joule per K
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d) 0.4 joule per K
Explanation
dU=µCvdT 80=5 ×Cv(120 - 100) Cv=4.0 joule/K Answer: (c)
Q.27
A Carnot's engine work as a refrigerator between 250K and 300K. If it receives 750 calories of heat from the reservoir at the lower temperature, the amount of heat rejected at the higher temperature is [ MPPMT 1990]
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a)900 calories
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b) 625 calories
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c)750 calories
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d)1000 calories
Explanation
A From the equations for efficiency of refrigerator Q1=Heat thrown to sinkQ2=Heat absorbedT1=Temperature of sourceT2=Temperature of sinkAnswer: (a)
Q.28
The work in an ideal monoatomic gas along the cyclic path LMNO is .. [ Raj. PMT 1996]
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a) PV
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b) 2PV
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c)3PV
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d)4PV
Explanation
Work=Area enclosed=PVAnswer: (a)
Q.29
During the adiabatic expansion of 2moles of a gas the internal energy of the gas is found to decrease by 2 joule. The work done during the process on gas will be equal to [ CPMT 1988]
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a)1 joule
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b) -1 joule
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c)2 joule
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d)-2 joule
Explanation
Decrease of internal energy means, the gas is expanding. Therefore the work done by the gas=2 jouleand the work done on the gas=-2J Answer:(d)
Q.30
The temperature of 5 moles of gas which was heated at constant volume was changed from 100°C to 120°C. The change in the internal energy of the gas was found to be 80 joule, the total heat capacity of the gas at constant volume will be equal to ..[ CPMT 1988]
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a) 8 joule per K
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b) 0.8 joule per K
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c) 4.0 joule per K
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d) 0.4 joule per K
Explanation
dU=µCvdT 80=5×Cv(120 - 100) Cv=0.8 joule per K Heat capacity = number of moles × CV Heat capacity = 5 × 0.8 = 4 joules per K Answer: (c)
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