Previous year questions (2014-20) - Thermodynamics Class 11 Notes | EduRev

Chemistry Class 11

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Q.1. An increase in the concentration of the reactants of a reaction leads to change in :     [2020]
(a) threshold energy
(b) collision frequency
(c) activation energy
(d) heat of reaction
Ans:
b
Solution: Collision frequency, as number of molecules per unit volume increases.

Q.2. The correct option for free expansion of an ideal gas under adiabatic condition is:     [2020]
(a) q < 0, ΔT = 0 and w = 0
(b) q > 0, ΔT > 0 and w > 0
(c) q = 0, ΔT = 0 and w = 0
(d) q = 0, ΔT < 0 and w > 0
Ans:
c
Solution: For an adiabatic process, q = 0
for free expansion, Pext = 0
∴ w = 0
& from First law of Thermodynamics, ΔE = 0
or ΔT = 0

Q.3. For the reaction, 2Cl(g) → Cl2(g), the correct option is:      [2020]
(a) ΔrH < 0 and ΔrS > 0
(b) ΔrH < 0 and ΔrS < 0
(c) ΔrH > 0 and ΔrS > 0
(d) ΔrH > 0 and ΔrS < 0
Ans: 
b
Solution:
2Cl(g) → Cl2(g) + Heat
Due to bond formation stability increases which results in release of heat.
∴ ΔHr = -ve or exothermic process
ΔHr < O
& ΔS < O, because number of Cl atoms decreases in the formation of Cl2(g)

Q.4. In which of the following processes, heat is neither absorbed nor released by a system?    [2019]
(a) Isothermal
(b) Adiabatic
(c) Isobaric
(d) Isochoric
Ans: 
b
Solution: 
In adiabatic process, there is no exchange of heat.

Q.5. A sample of 0.1 g of water at 100°C and normal pressure (1.013 x 105 Nm–2) requires 54 cal of heat energy to convert to steam at 100°C. If the volume of the steam produced is 167.1 cc, the change in internal energy of the sample, is:-    [2018]
(a) 104.3 J
(b) 208.7 J
(c) 42.2 J
(d) 84.5 J
Ans: 
c
Solution: ΔQ = 54 cal = 54 × 4.18 joule = 225.72 joule
ΔW = P[Vsteam – Vwater] [For water 0.1 gram=0.1 cc]
= 1.013 × 105[167.1 × 10–6 – 0.1 × 10–6] joule
= 1.013 × 167 × 10–1 = 16.917 joule
By FLOT
ΔU = ΔQ – ΔW = 225.72 – 16.917
ΔU = 208.8 joule

Q.6. The volume (V) of a monatomic gas varies with its temperature (T), as shown in the graph. The ratio of work done by the gas, to the heat absorbed by it, when it undergoes a change from state A to state B, is:-    [2018]
Previous year questions (2014-20) - Thermodynamics Class 11 Notes | EduRev
(a) 2/5
(b) 2/3
(c) 1/3
(d) 2/7
Ans:
a
Solution:
Previous year questions (2014-20) - Thermodynamics Class 11 Notes | EduRev

Q.7. The efficiency of an ideal heat engine working between the freezing point and boiling point of water, is:-    [2018]
(a) 26.8%
(b) 20%
(c) 6.25%
(d) 12.5%
Ans: 
a
Solution:
Previous year questions (2014-20) - Thermodynamics Class 11 Notes | EduRev

Q.8. Thermodynamic processes are indicated in the following diagram:    [2017]
Previous year questions (2014-20) - Thermodynamics Class 11 Notes | EduRev
Previous year questions (2014-20) - Thermodynamics Class 11 Notes | EduRev
(a) P → c, Q → a, R → d, S → b
(b) P → c, Q → d, R → b, S → a
(c) P → d, Q → b, R → a, S → c
(d) P → a, Q → c, R → d, S → b
Ans:
a
Solution:
Process (1) → volume constant → Isochroic
Process (2) → adiabatic
Process (3) → Temperature constant → Isothermal
Process (4) → Pressure constant → Isobaric

Q.9. A carnot engine having an efficiency of 1/2 as heat engine, is used as a refrigerator. If the work done on the system is 10 J, the amount of energy absorbed from the reservoir at lower temperature is:-    [2017]
(a) 90 J
(b) 99 J
(c) 100 J
(d) 1 J
Ans:
a
Solution:
Previous year questions (2014-20) - Thermodynamics Class 11 Notes | EduRev

Q.10. A refrigerator works between 40 celcius and 300 celcius. 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)    [2016]
(a) 2365W
(b) 2.365W
(c) 23.65W
(d) 236.5W
Ans:
d
Solution: 
Given,
Temperature of the source, T = 30o C = 30 + 273  = 303 K
Temperature of sink, T2 = 4o C = 4 + 273 = 277 K
We know,
Previous year questions (2014-20) - Thermodynamics Class 11 Notes | EduRev
where, Q2 is the amount of heat drawn from the sink at T2,
W is the work done on the working substance,
Q1 is the amount of heat rejected to source at room temperature T1.
That is,
Previous year questions (2014-20) - Thermodynamics Class 11 Notes | EduRev
Previous year questions (2014-20) - Thermodynamics Class 11 Notes | EduRev

Q.11. A gas is compressed isothermally to half its initial volume. The same gas is compressed separately through an adiabatic process until its volume is again reduced to half. Then    [2016]
(a) Which of the case (whether compression through isothermal or through adiabatic process) requires more work will depend upon the atomicity of the gas
(b) Compressing the gas isothermally will require more work to be done
(c) Compressing the gas through adiabatic process will require more work to be done
(d) Compressing the gas isothermally or adiabatically will require the same amount of work
Ans:
c
Solution: Directly from graph the magnitude of work done = Area under p-v plot is larger for adiabatic compression
Previous year questions (2014-20) - Thermodynamics Class 11 Notes | EduRev

Q.12. One mole of an ideal diatomic gas undergoes a transition from A to B along a path AB as shown in the figure,
Previous year questions (2014-20) - Thermodynamics Class 11 Notes | EduRev
The change in internal energy of the gas during the transition is :    [2015]
(a) -12 kJ
(b) 20 KJ
(c) - 20 KJ
(d) 20 J
Ans:
c
Solution:
Previous year questions (2014-20) - Thermodynamics Class 11 Notes | EduRev

Q.13. A Carnot engine, having an efficiency of η = 1/10 as heat engine, is used as a refrigerator. If the work done on the system is 10 J, the amount of energy absorbed from the reservoir at lower temperature is:    [2015]
(a) 1 J
(b) 100 J
(c) 99 J
(d) 90 J
Ans: d

Solution:
Previous year questions (2014-20) - Thermodynamics Class 11 Notes | EduRev

Q.14. Figure below shows two paths that may be taken by a gas to go from a state A to a state C.
Previous year questions (2014-20) - Thermodynamics Class 11 Notes | EduRev
In process AB, 400 J of heat is added to the system and in process BC, 100 J of heat is added to the system. The heat absorbed by the system in the process AC will be:    [2015]
(a) 300 J
(b) 380 J
(c) 500 J
(d) 460 J
Ans:
d
Solution:
Previous year questions (2014-20) - Thermodynamics Class 11 Notes | EduRev
See figure alongside
Process AB is isochoric so no work is done.
Heat added to be system is Q = 400 J.
Q = ΔU + ΔW
where ΔU is the change in internal energy
ΔW is the work done.
Since ΔW = 0
ΔU = Q = 400 J
Change in internal energy is 400 J.
Process BC is isobaric and the work done is given by
ΔW = P(V2 -V1) = 6 x 104(4 x 10-3 - 2 x 10-3)
= 6 x 104 x 2 x 10-3 = 120 J
Heat added to be system is Q = 100 J.
Since Q = ΔU + ΔW
 ∴ ΔU = Q- ΔW = (100 - 120) J = -20 J
Change in internal energy is - 20 J.Total increase in internal energy is going from state A to state C is 400 - 20 = 380 J
Work done in process AC is the area under the curve Area of the trapezium
Previous year questions (2014-20) - Thermodynamics Class 11 Notes | EduRevSince Q = ΔU + ΔW
and ΔU the change in internal energy in process AC, we have
ΔU = 380 J and ΔW = 80 J
∴ Q = ΔU + ΔU = 380 + 80 = 460 J

Q.15. A monoatomic gas at a pressure P, having a volume V expands isothermally to a volume 2v and then adiabatically to a volume 16v. The final pressure of the gas is: (Take γ = 5/3)    [2014]
(a) P/64
(b) 16P
(c) 64P
(d) 32P
Ans:
a
Solution:
Previous year questions (2014-20) - Thermodynamics Class 11 Notes | EduRev
Previous year questions (2014-20) - Thermodynamics Class 11 Notes | EduRev

Q.16. A thermodynamics system undergoes cyclic process ABCDA as shown in fig. The work done by the system in the cycle is:    [2014]
Previous year questions (2014-20) - Thermodynamics Class 11 Notes | EduRev
(a) P0V0/2
(b) zero
(d)  2P0V0
Ans:
b
Solution:
Previous year questions (2014-20) - Thermodynamics Class 11 Notes | EduRev

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