Thermodynamics Test

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Thermodynamics Test - 5

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Question 1
4 / -1

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
4 / -1

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
4 / -1

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
4 / -1

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
4 / -1

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
4 / -1

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
4 / -1

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
4 / -1

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
4 / -1

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
4 / -1

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
Question 11
4 / -1

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]

A

B

C

D

Solution

∆S_X→Z = ∆S_X→Y + ∆S_Y→Z(Entropy is a state function, so it is additive)
W_X→Y→Z = WX→Y (work done in _y→z is zero as the process is isochoric)
Question 12
4 / -1

Identify the correct statement regarding entropy:

A
At absolute zero temperature, entropy of a perfectly crystalline substance is taken to be zero

B
At absolute zero of temperature the entropy of a perfectly crystalline substance is +ve

C
At absolute zero of temperature the entropy of all crystalline substance is to be zero

D
At 0^∘C, the entropy of a perfectly crystalline substance is taken to be zero

Solution

At absolute zero temperature, entropy of a perfectly crystalline substance is taken to be zero. It is called third law of thermodynamics.
Question 13
4 / -1

In an endothermic reaction, the value of ΔH is:

A
zero

B
positive

C
negative

D
constant

Solution

For endothermic, reactions standard heat of reaction (ΔH) is positive because in these total energy of reactant is lower than product. i.e.
Question 14
4 / -1

In which of the following cases, entropy of I is larger than that of II?

A
a

B
b

C
c

D
d

Solution

a) More molar mass, more entropy. So ∆_SN2O4 > ∆S_NO2
b) CO₂ has more entropy than dry ice at -78°C
c) Pure alumina iss crystalline solid while ruby is amorphous. And ∆S_amorphous > ∆S_Crystalline. So alumina has less entropy than ruby.
d) At lower pressure, entropy be higher as gas particles are far from each other. So (∆S_N2)1_bar > (∆S_N2)_{5 bar}
Question 15
4 / -1

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.

A

B

C

D
All of these

Solution

AB, CD and EF are isotherms.
AC, CE, BD and D F are adiabates.
For adiabatic process, we have change n entropy = 0
So Entropy_A→B = Entropy_C→D = Entropy_E→F and Entropy_A→C = Entropy_A→C = Entropy_C→E = Entropy_F→D = Entropy_D→B
By this, all options are correct.
Question 16
4 / -1

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

A

B

C

D
Question 17
4 / -1

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

A
6.00 bar

B
4.00 bar

C
2.00 bar

D
2.67 bar

Solution

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
Question 18
4 / -1

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

A
0.96 JK^-1

B
3.35 JK^-1

C
- 2.40 JK^-1

D
- 0.96 JK^-1
Question 19
4 / -1

2 mole of ideal gas at 27^∘C temperature is expanded reversibly from 2 L to 20 L. Find entropy change (R = 2 cal/mol K):

A
92.1

B
0

C
4

D
9.2

Solution
Question 20
4 / -1

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?

A
100 %

B
80 %

C
75 %

D
70 %

Solution

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,

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