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Physics NEET MCQ
Heat And Thermodynamics Mcq
Quiz 3
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Q.1
In the given pressure-volume diagram, the isochoric, isothermal, isobaric and isotropic parts respectively are .. [ UCET 1995]
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a)BA, AD, DC, CB
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b) DC, CB, BA, AD
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c)AB, BC, CD, DA
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d)CD, DA, AB, BC
Explanation
Answer: (d)
Q.2
In a thermodynamic process, pressure of a fixed mass of a gas is changed in such a manner that the gas molecules gives out 30 joule of heat and 10 joule of work is done on the gas. If the initial internal energy of the gas was 40 joule, then the final internal energy will be ..[ CPMT 1988]
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a) zero
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b) 80 joule
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c)20 joule
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d)-20 joule
Explanation
dQ=-30J ( negative indicates heat is going out of the system)W=-10J Ui=40J dQ=dU + W-30=(Uf -40) +(-10) Uf=20JAnswer: (c)
Q.3
When water is heated from 0° to 4° then ..[ AP 1986]
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a) Cp=Cv
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b) Cp > Cv
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c) Cp < Cv
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d) Cp - Cv=R
Explanation
Cp < Cv, because on heating upto 4°C, water contracts in volume Answer:(c)
Q.4
One mole of an ideal mono atomic gas is heated at a constant pressure of one atmosphere from 0° to 100°. Then work done by the gas is [ CPMT 1991]
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a) 6.56 joules
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b) 8.32×102 joules
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c) 12.48×102 joules
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d) 20.8×102 joules
Explanation
for monoatomic gas degree of freedom=3 thus Cv=3/2 and Cp=5/2 dQ=µCpdT dQ=1×(5/2)R×100 dQ=2077.5J=20.8 ×102J dU=µCvdT=1×(3/2)R×100dU=1246.5JW=dQ - dU=8.31×102J Answer: (b)
Q.5
The curve which represents an adiabatic change is [ AMU 1995]
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a)
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b)
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c)
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d)
Explanation
Answer: (c)
Q.6
The amount of work done in a an adiabatic expansion from temperature T to T' for a gram mole of gas is given by
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a) R (T - T')
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b) R(T - T') / (γ- 1)
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c)RT
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d)R(T-T') (γ - 1)
Explanation
Answer: (b)
Q.7
A monoatomic gas initially at 180°C is compressed adiabatically to one-eighth of its original volume. The temperature after compression will be [ CBSE 1996]
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a) 180°C
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b) 144°C
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c)891°C
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d)887°C
Explanation
For adiabatic process T1V1γ-1=T2V2γ-1 for mono atomic gas γ=5/3 and T1=18+273=291K Answer:(c)
Q.8
At same temperature, pressure and volume the quantity which is same between gases will be [ Raj.PMT 1996]
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a) Number of atoms
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b) Average kinetic energy
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c) Mean square velocity
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d) Mean free path
Explanation
Answer: (a)
Q.9
In isothermic process, which statement is wrong
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a)Temperature is constant
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b) Internal energy will be constant
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c)No exchange of energy
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d)(a) and (b) are correct
Explanation
Answer: (c)
Q.10
One mole of an ideal mono-atomic gas changes from state A to B as shown in figure. In this process the work done and change in internal energy are
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a) P1V1 ; 3P1V1
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b) (3/2)P1V1 ; (9/2)P1V1
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c)(3/2)P1V1 ; (3)P1V1
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d)(3/2)P1V1 ; P1V1
Explanation
W=average pressure × Change in volume From gas law T1=(P1 V1) / R andT2=(2P1 × 2V1 )/ R ∴ T2 - T1=3P1 × V1 / R Now Answer: (b)
Q.11
For adiabatic process, wrong statement is [ Raj. PMT 1997]
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a) dQ=0
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b) dU - -W
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c)Q=constant
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d)Entropy is not constant
Explanation
Answer:(d)
Q.12
The process in which no heat enters or leaves the system is termed as [ BHU 1998]
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a) isochoric
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b) isobaric
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c) isothermal
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d) adiabatic
Explanation
Answer: (d)
Q.13
Two samples A and B of a gas initially at the same temperature and pressure, are compressed from volume V to V/2 ( A isothermally and b adiabatically). The final pressure of [ MPPMT 1997]
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a)A is greater than that of B
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b) A is equal to that of B
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c)A is less than that of B
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d)A is twice that of B
Explanation
Answer: (c)
Q.14
If one mole of monoatomic gas ( γ=5/3) is mixed with one mole of diatomic gas ( γ=7/5). The value of γ for the mixture is [ Raj.PET 1996]
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a) 1.4
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b) 1.5
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c)1.53
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d)3.07
Explanation
for mono atomic gas Cp=3/2for diatomic gas Cp=7/2 dQ1=1 × (5R/2) ×dTdQ2=1× (7R/2) ×dT ∴ dQ=dQ1 + dQ2 dQ=6RdTNow there are to moles of gas on mixing 2×Cp ×dT=6RdTOr Cp=3R and Cv=2Rγ=Cp /Cv=3/2=1.5Answer: (b)
Q.15
In an ideal gas, the energy is .. [ AFMC 1994]
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a) wholly kinetic
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b) wholly potential
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c)sum of the two
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d)all are true
Explanation
Answer:(a)
Q.16
Arrange in ascending order of work done [ Raj. PMT 1996]
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a) Adiabatic > Isothermal > Isobaric
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b) Isothermal >Adiabatic > Isochoric
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c) Adiabatic > Isobaric > Isothermal
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d) None of the above
Explanation
Work = area under the curve and from figure it can be seen that, isobaric > isothermal > adiabatic > isochoric Answer: (b)
Q.17
The isothermal bulk modulus of an ideal gas at pressure P is
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a)P
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b) γP
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c)P/2
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d) P / γ
Explanation
Answer: (a)
Q.18
The adiabatic bulk modulus of a ideal gas at pressure P is given by
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a) P
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b) 2P
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c)P/2
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d)γP
Explanation
Answer: (d)
Q.19
Half mole of an ideal gas at temperature of 27°C undergoes an isothermal expansion so that the final volume is double the initial volume. The work done by the gas is ( R=8.3 J/mol K)
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a) 860 J
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b) 1720 J
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c)0.78J
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d)zero
Explanation
Answer:(a)
Q.20
The adiabatic elasticity of hydrogen gas at N.T.P is ( γ=1.4) [ MPPMT 1990]
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a) 1×105 N/m2
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b) 1×10-8 N/m2
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c) 1.4 N/m2
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d) 1.4×105 N/m2
Explanation
Elasticity=bulk modulus=γP Elasticity=1.4 × 105 N/m2 Answer: (d)
Q.21
For ideal gas , which statement is not true.. [ Raj. PMT 1997]
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a)It obeys Boyle's law
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b) Follows PV=RT
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c)Internal energy depends on temperature only
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d)Follows Vander Waal's equation
Explanation
Answer: (d)
Q.22
An ideal gas at 27°C is compressed adiabatically to 8/27 of its original volume [ TVγ-1=constant and γ=5/3]. Then the rise in temperature is [ CPMT 1998]
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a) 450°C
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b) 375°C
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c)225°C
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d)400°C
Explanation
Thus rise in temperature=675 -300=375K=375°CAnswer: (b)
Q.23
If R is the gas constant for 1 gm mole, and Cp and Cv are specific heats for solid per mole at constant pressure and volume respectively, then
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a)Cp - Cv=R
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b) Cp - Cv < R
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c)Cp - Cv=0
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d)Cp - Cv > R
Explanation
Take the equation, U = q + P Δ(V) , for constant volume , Δ(V) ofcourse is 0, because their is no change in the volume ( since, it is at constant volume). Thus we get , U = q + P (0) ; U = q. That is Cv And for constant pressure, U = q + P del(V) , but for solids and liquids , change is volume is nearly 0 but not zero or Cp ≠q Thus for pratical purpose Cp = Cv but Theoritically Cp≠ Cv, ∴ Option "b" is correct Answer:(b)
Q.24
If for hydrogen Cp - Cv=a for oxygen Cp - Cv=b where Cp and Cv refers to specific heat at constant pressure and constant volume per unit mass, the [ CBSE 1993]
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a) a=b
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b) a=16 b
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c) b=16a
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d) a and b are not related
Explanation
Specific heat are per unit mass and not per mole hence should be converted per mole, Given data if for Specific heat of gas converting equation to molar specific heat we must multiply given equation by molecular weight M Thus molar specific heat difference M[Cp - Cv]=Ma=RFor Hydrogen 2[Cp - Cv]=2a=RFor Oxygen 32[Cp - Cv]=32b=R 2a=32 b ∴ a=16bAnswer: (b)
Q.25
A gas expands from a volume of 2 m3 to 6m3 at a constant pressure of 10N/m2 and then at a constant volume of 6m3, the pressure is changed from 10N/m2 to 20N/mThe work done by the gas is [ CBSE 1993]
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a)100 J
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b) 40 J
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c)40 erg
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d)60 J
Explanation
Here pressure and volume has changed Answer: (d)
Q.26
A nonconducting partition divides a container into two equal compartments. One filled with helium gas at 200K and the other is filled with oxygen gas at 400k. The number of molecules in each gas is the same. If the partition is removed to allow the gases to mix, the final temperature will be [ CBSE 1993]
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a) 350 K
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b) 325 K
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c)300 K
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d)275 K
Explanation
For Helium P1V1=nRT1For Oxygen gasP2V2=nRT2After removing partition gases gets mixedP1V1 + P2V2=PV∴ nkT1 + nkT2=PVBut after mixing PV=2nKT ∴ nkT1 + nkT2=2nkT T1 + T2=2T 200 +400=2T T=300KAnswer: (c)
Q.27
Work done by 0.1 mole of gas at 27°C to double its volume at constant pressure is ..(R=2 cal/mol K) [ EAMCET 1994]
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a) 54 cal
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b) 600 cal
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c)60 cal
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d)546 cal
Explanation
W=PΔV=PVBut PV=µRT W=0.1×2×300=60cal Answer:(c)
Q.28
Which is correct statement? [ MPPMT 1993]
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a) For isothermal change PV=constant
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b) In an isothermal process the change in internal energy must be equal to the work done
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c) For an adiabatic change P2 / P1=(V2 / V1)γ, where γ is the ratio of specific heats
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d) In an adiabatic process, the external work done must be equal to the heat entering the system
Explanation
Answer: (a)
Q.29
A body of mass 5kg falls from a height of 20m on the ground and it rebounds to a height of 0.2m. If the loss in potential energy is used up by the body, then what will be the temperature rise? ( specific heat of material=0.09 cal/gm°C) .. [ CPMT 1998]
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a)5°C
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b) 4°C
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c)8°C
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d)none of above
Explanation
heat produced=loss in potential energy heat=mg ( h-h' ) heat=5×10(20-0.2)=990 jouleheat=990 /4.2 calThis energy is used to increase temperature of the bodyQ=msΔT 990 /4.2=5000 × 0.09 ×ΔT ΔT=11/21 °CAnswer: (d)
Q.30
The density of steam is 0.6 kg/m3, 1gm water at 100°C and 1×105 N/m2 pressure is converted into steam. The external work done nearly
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a) 0.6×105 J
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b) 1700 J
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c)17 J
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d)170 J
Explanation
density of steam 0.6 kg/m3=0.6×10-3 g/cc Volume of steam=1 / (0.6×10-3 )=1666.66 cc=1666.66×10-6 m3 Volume of 1 cc water=10-6 m3 W=pΔP W=105 [ 1667 - 1]×10-6 W=166.7 J=170 JAnswer: (d)
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