Thermodynamics Test

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Thermodynamics Test - 20

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Weekly Quiz Competition

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)(ΔH_vap = 40850 J mol^-1)

A
+109.52 JK^-1 mol^-1

B
-109.52 JK^-1 mol^-1

C
0.00 JK^-1 mol^-1

D
+6.084 JK^-1 mol^-1

Solution

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

A
5705.8 JK^-1

B
19.15 JK^-1

C
38.29 JK^-1

D
-19.15 JK^-1

Solution

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

A
-22.98 JK^-1

B
22.98 JK^-1

C
57.4 JK^-1

D
80.42 JK^-1

Solution

∆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
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

A
2.422 JK^-1

B
5.98 JK^-1

C
- 2.422 JK^-1

D
- 5.981 JK^-1

Solution

For isobaric process, we have ∆S =nC_pln(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 =nC_pln(T₂/T₁)
∆S =0.40×5/2R×ln(400/300) = 2.38 JK^-1
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

A
0.335 JK^-1

B
-0.33 JK^-1

C
0.670 JK^-1

D
0.00 JK^-1
Question 6
1 / -0

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

A
10.07 cal K^-1

B
2.013 kcal K^-1

C
305.4 cal K^-1

D
- 2.013 kcal K^-1

Solution

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
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

A
1.474 bar

B
0.678 bar

C
0.796 bar

D
1.00 bar

Solution

Isentropic process means that entropy is constant. This is true only for reversible adiabatic process.
Applying P₁^1-γ T₁^γ = P₂^1-γ T₂^γ (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
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

A
AB, BC

B
BC,CD

C
BC, DE

D
CD, DE

Solution

You can co-relate both graph and the result will be option c.
Question 9
1 / -0

ΔH_vap = 30 kJ mol^-1 and ΔS_vap = 75 J mol^-1 K^-1. Thus, temperature of the vapour at 1 atm is
[IIT JEE 2004]

A
400 K

B
350 K

C
298 K

D
250 K

Solution

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
Question 10
1 / -0

For the process, and 1 atmosphere pressure, the correct choice is
[JEE Advanced 2014]

A
ΔS_(System) > 0, ΔS_{(surrounding)}> 0

B
ΔS_(System) > 0, ΔS_{(surrounding)}< 0

C
ΔS_(System) < 0, ΔS_{(surrounding)}> 0

D
ΔS_(System) < 0, ΔS_{(surrounding)} < 0

Solution

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

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Thermodynamics Test - 20

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Thermodynamics Test - 20

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Question 1

+109.52 JK-1 mol-1

-109.52 JK-1 mol-1

0.00 JK-1 mol-1

+6.084 JK-1 mol-1

Solution

Question 2

5705.8 JK-1

19.15 JK-1

38.29 JK-1

-19.15 JK-1

Solution

Question 3

-22.98 JK-1

22.98 JK-1

57.4 JK-1

80.42 JK-1

Solution

Question 4

2.422 JK-1

5.98 JK-1

- 2.422 JK-1

- 5.981 JK-1

Solution

Question 5

0.335 JK-1

-0.33 JK-1

0.670 JK-1

0.00 JK-1

Question 6

10.07 cal K-1

2.013 kcal K-1

305.4 cal K-1

- 2.013 kcal K-1

Solution

Question 7

1.474 bar

0.678 bar

0.796 bar

1.00 bar

Solution

Question 8

AB, BC

BC,CD

BC, DE

CD, DE

Solution

Question 9

400 K

350 K

298 K

250 K

Solution

Question 10

ΔS(System) > 0, ΔS(surrounding) > 0

ΔS(System) > 0, ΔS(surrounding) < 0

ΔS(System) < 0, ΔS(surrounding) > 0

ΔS(System) < 0, ΔS(surrounding) < 0

Solution

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+109.52 JK^-1 mol^-1

-109.52 JK^-1 mol^-1

0.00 JK^-1 mol^-1

+6.084 JK^-1 mol^-1

5705.8 JK^-1

19.15 JK^-1

38.29 JK^-1

-19.15 JK^-1

-22.98 JK^-1

22.98 JK^-1

57.4 JK^-1

80.42 JK^-1

2.422 JK^-1

5.98 JK^-1

- 2.422 JK^-1

- 5.981 JK^-1

0.335 JK^-1

-0.33 JK^-1

0.670 JK^-1

0.00 JK^-1

10.07 cal K^-1

2.013 kcal K^-1

305.4 cal K^-1

- 2.013 kcal K^-1

ΔS_(System) > 0, ΔS_{(surrounding)}> 0

ΔS_(System) > 0, ΔS_{(surrounding)}< 0

ΔS_(System) < 0, ΔS_{(surrounding)}> 0

ΔS_(System) < 0, ΔS_{(surrounding)} < 0