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Test: Second Law of Thermodynamics & Entropy - NEET MCQ


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20 Questions MCQ Test - Test: Second Law of Thermodynamics & Entropy

Test: Second Law of Thermodynamics & Entropy for NEET 2024 is part of NEET preparation. The Test: Second Law of Thermodynamics & Entropy questions and answers have been prepared according to the NEET exam syllabus.The Test: Second Law of Thermodynamics & Entropy MCQs are made for NEET 2024 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for Test: Second Law of Thermodynamics & Entropy below.
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Test: Second Law of Thermodynamics & Entropy - Question 1
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Direction (Q. Nos. 1-10) This section contains 10 multiple choice questions. Each question has four choices (a), (b), (c) and (d), out of which ONLY ONE option is correct.

Q. Entropy change for the following reversible process is 1 mole H2O

(l, 1 atm, 100°C ) 1 mole H2O (g , 1 atm, 100°C)(ΔHvap = 40850 J mol-1)

Detailed Solution for Test: Second Law of Thermodynamics & Entropy - Question 1

∆S=Change in energy/Absolute temp =∆H/(100+273)K =40850/373 = 109.52 J/K/mol

Test: Second Law of Thermodynamics & Entropy - Question 2
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Entropy change when 2 moles of an ideal gas expands reversibly from an initial volume of 1 dm3 to a final volume of 10 dm3 at a constant temperature of 298 K is

Detailed Solution for Test: Second Law of Thermodynamics & Entropy - Question 2

∆S= 2.303nR × log(V2/V1)Here n=2, R=8.314, V2= 10, V1= 1

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Test: Second Law of Thermodynamics & Entropy - Question 3
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3 moles of a diatomic gas are heated from 127° C to 727° C at a constant pressure of 1 atm. Entropy change is (log 2.5 = 0 .4)

Detailed Solution for Test: Second Law of Thermodynamics & Entropy - Question 3

∆S = nCplnT2/T1 + nRlnP1/P2
Since pressure is constant, so the second term will be zero.
Or ∆S = 3×7/2×8.314×2.303×log(1000/400)
= 80.42 JK-1

Test: Second Law of Thermodynamics & Entropy - Question 4
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10 dm3 of an ideal monoatomic gas at 27° C and 1.01 x 105 Nm-2 pressure are heated at constant pressure to 127°C. Thus entropy change is
Detailed Solution for Test: Second Law of Thermodynamics & Entropy - Question 4

For isobaric process, we have ∆S =nCpln(T2/T1)
T2 = 273+127 = 400K and T1 = 273+227 = 300K
Applying pV = nRT at initial condition,
1×10 = n×0.0821×300
n = 0.40
Applying ∆S =nCpln(T2/T1)
∆S =0.40×5/2R×ln(400/300) = 2.38 JK-1

Test: Second Law of Thermodynamics & Entropy - Question 5
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Exactly 100 J of heat was transferred reversibly to a block of gold at 25.00° C from a thermal reservoir at 25.01 °C, and then exactly 100 J of heat was absorbed reversibled from the block of gold by a thermal reservoir at 24.99° C. Thus entropy change of the system is
Test: Second Law of Thermodynamics & Entropy - Question 6
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Given

I. C (diamond) + O2(g) → CO2(g) ; ΔH° = - 91.0 kcdl mol-1
II. C(graphite) + O2(g) → CO2(g) ; ΔH° = - 94.0 kcal mol-1

Q. At 298 K, 2.4 kg of carbon (diamond) is converted into graphite form. Thus, entropy change is
Detailed Solution for Test: Second Law of Thermodynamics & Entropy - Question 6

The reaction is 
C(diamond)     →     C(graphite)      ∆H = (94-91) = 3 kcal mol-1
∆S = ∆H/T
∆H = (94-91)×2.4×103/12   
= 600 kcal
∆S = 600/298 = 2.013 kcal K-1

Test: Second Law of Thermodynamics & Entropy - Question 7
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Consider a reversible isentropic expansion of 1.0 mole of an ideal monoatomic gas from 25°C to 75°C. If the initial pressure was 1.0 bar, final pressure is 
Detailed Solution for Test: Second Law of Thermodynamics & Entropy - Question 7

Isentropic process means that entropy is constant. This is true only for reversible adiabatic process.
Applying P11-γ T1γ = P21-γ T2γ (for monatomic species, γ = 5/3)
(1/P)-⅔ = (75+273/25+273)5/3
Or (1/P)-2 = (348/298)5
Or P = 1.474 bar

Test: Second Law of Thermodynamics & Entropy - Question 8
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Consider the following figure representing the increase in entropy of a substance from absolute zero to its gaseous state at some temperature

Q. ΔS° (fusion) and ΔS° (vaporisation) are respectively indicated by 
Detailed Solution for Test: Second Law of Thermodynamics & Entropy - Question 8


You  can co-relate both graph and the result will be option c.

Test: Second Law of Thermodynamics & Entropy - Question 9
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ΔHvap = 30 kJ mol-1 and ΔSvap = 75 J mol-1 K-1. Thus, temperature of the vapour at 1 atm is

[IIT JEE 2004]
Detailed Solution for Test: Second Law of Thermodynamics & Entropy - Question 9

Vapour pressure is equal to atmospheric pressure ,it means the substance is at boiling point
At boiling point, liquid and Gas are in equilibrium. Therefore dG=0
dG = H - TdS
dG = 0
⇒H = TdS
⇒T = H/dS
⇒ T = 30 103/75 = 400K

Test: Second Law of Thermodynamics & Entropy - Question 10
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For the process, and 1 atmosphere pressure, the correct choice is

[JEE Advanced 2014]
Detailed Solution for Test: Second Law of Thermodynamics & Entropy - Question 10

At 100°C and 1 atmosphere pressure H2O (l) ⇋ H2O(g) is at equilibrium. For equilibrium

*Multiple options can be correct
Test: Second Law of Thermodynamics & Entropy - Question 11
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Direction (Q. Nos. 11-14) This section contains 4 multiple choice questions. Each question has four choices (a), (b), (c) and (d), out of which ONE or  MORE THANT ONE  is correct.

Q. For an ideal gas, consider only (p -V) work in going from initial state X to the final state Z. The final state Z can be reached either of the two paths shown in the figure. Which of the following choice (s) is (are) correct?

(Take ΔS as change in entropy and W as work done)

[IIT JEE 2012]
Detailed Solution for Test: Second Law of Thermodynamics & Entropy - Question 11

∆SX→Z = ∆SX→Y + ∆SY→Z(Entropy is a state function, so it is additive)
WX→Y→Z = WX→Y (work done in y→z is zero as the process is isochoric)

*Multiple options can be correct
Test: Second Law of Thermodynamics & Entropy - Question 12
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Benzene and naphthalene form an ideal solution at room temperature. For this process, the true statement(s) is (are)

[JEE Advanced 2013]
Detailed Solution for Test: Second Law of Thermodynamics & Entropy - Question 12

When an ideal solution is formed process is spontaneous

According to Raoult's law, for an ideal solution

ΔH=0

ΔVmix​=0

Since there is no exchange of heat energy between system and surroundings

ΔSsurroundings​=0

ΔSsys.​=+ve

∴ From the relation

ΔG=ΔH–TΔS

ΔG=−ve

Hence, the correct options are B, C and D

*Multiple options can be correct
Test: Second Law of Thermodynamics & Entropy - Question 13
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Consider the following process

Select correct choices(s)
Detailed Solution for Test: Second Law of Thermodynamics & Entropy - Question 13

∆SX→Y = ∆SX→Z + ∆SZ→Y(Entropy is a state function, so it is additive)
However, ∆SX→Z is zero. As for adiabatic process. There is no change in entropy. 
∆SX→Y =  ∆SZ→Y = 2.303 5 8.314 log(10/1) = 95.7 JK-1

*Multiple options can be correct
Test: Second Law of Thermodynamics & Entropy - Question 14
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In which of the following cases, entropy of I is larger than that of II?
Detailed Solution for Test: Second Law of Thermodynamics & Entropy - Question 14

a) More molar mass, more entropy. So ∆SN2O4 > ∆SNO2
b) CO2 has more entropy than dry ice at -78°C
c) Pure alumina iss crystalline solid while ruby is amorphous. And ∆Samorphous > ∆SCrystalline. So alumina has less entropy than ruby.
d) At lower pressure, entropy be higher as gas particles are far from each other. So (∆SN2)1 bar > (∆SN2)5 bar 
 

Test: Second Law of Thermodynamics & Entropy - Question 15
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Direction (Q. Nos. 15 and 18) This section contains 2 paragraphs, each describing theory, experiments, data etc. Four questions related to the paragraphs have been given. Each question has only one correct answer among the four given options (a), (b), (c) and (d)

Passage I

Consider a series of isotherms and adiabates as shown

AB, CD and EF are isotherms.
AC, CE, BD and D F are adiabates.

 

Q.

Detailed Solution for Test: Second Law of Thermodynamics & Entropy - Question 15


AB, CD and EF are isotherms.
AC, CE, BD and D F are adiabates. 
For adiabatic process, we have change n entropy = 0
So EntropyA→B = EntropyC→D = EntropyE→F and EntropyA→C = EntropyA→C   = EntropyC→E = EntropyF→D = EntropyD→B
By this, all options are correct.  

Test: Second Law of Thermodynamics & Entropy - Question 16
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Passage I

Consider a series of isotherms and adiabates as shown

AB, CD and EF are isotherms.
AC, CE, BD and D F are adiabates.

 

Q. Select incorrect relationship

 
Test: Second Law of Thermodynamics & Entropy - Question 17
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Passage II

The stopcock connecting A and B is of negligible volume. Stopcock is opened and gases are allowed to mix isothermally.

 

Q. Final pressure set up is
Detailed Solution for Test: Second Law of Thermodynamics & Entropy - Question 17


Volume in (I) = nRT/P1 = 1× 0.0821× 298/4 = 6.11 L
Volume in (II) = nRT/P2 = 1× 0.0821× 298/2 = 12.23 L
Total volume = 18.34 L
Applying PV = nRT at final condition, 
P = 2× 0.0821× 298/18.34 = 2.66

Test: Second Law of Thermodynamics & Entropy - Question 18
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Passage II

The stopcock connecting A and B is of negligible volume. Stopcock is opened and gases are allowed to mix isothermally.

 

Q. Entropy change for the system is 
*Answer can only contain numeric values
Test: Second Law of Thermodynamics & Entropy - Question 19
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Direction (Q. Nos. 19 and 20) This section contains 2 questions. Each question, when worked out will result in an integer from 0 fo 9 (both inclusive).

Q. Consider two Carnot engines (1) and (2)

Efficiency η2
Derive the value of η2  η1

Detailed Solution for Test: Second Law of Thermodynamics & Entropy - Question 19

η1 = 1-273/298 = 25/298
η2 = 1-248/298 = 50/298
η2/η1 = (50/298)/25/298 = 2

Test: Second Law of Thermodynamics & Entropy - Question 20
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A cyclic heat engine operates between a source temperature of 927 oC and a sink temperature of 27 oC. What will be the maximum efficiency of the heat engine?
Detailed Solution for Test: Second Law of Thermodynamics & Entropy - Question 20

Heat engine is operated between the temperatures

T1 = 927 0C = 927 + 273 = 1200 K and
T2 = 27 0C = 27 + 273 = 300 K

The maximum efficiency of a heat engine is given by,

ηmax = 1 – (T2/T1)
ηmax = 1 – (300/1200)
ηmax = 0.75

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