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Chapter 6- Thermodynamics Interview Questions Answers

Question 1 :
Choose the correct answer. A thermodynamic state function is a quantity
(i) used to determine heat changes
(ii) whose value is independent of path
(iii) used to determine pressure volume work
(iv) whose value depends on temperature only.

Answer 1 :

A thermodynamic state function is a quantity whose value is independent of a path.
Functions like p, V, T etc. depend only on the state of a system and not on the path.
Hence, alternative (ii) is correct.

Question 2 :

For the process to occurunder adiabatic conditions, the correct condition is:

(i) Δ= 0

(ii) Δ= 0

(iii) = 0

(iv) w = 0

Answer 2 :

A system is said to be under adiabaticconditions if there is no exchange of heat between the system and itssurroundings. Hence, under adiabatic conditions, = 0.

Therefore,alternative (iii) is correct.

Question 3 :

The enthalpies of allelements in their standard states are:

(i) unity

(ii) zero

(iii) < 0

(iv)different for each element

Answer 3 :

The enthalpy of allelements in their standard state is zero.

Therefore,alternative (ii) is correct.

Question 4 :

ΔUθof combustion of methaneis – X kJ mol–1. The value of ΔHθ is

(i) = ΔUθ

(ii) > ΔUθ

(iii) < ΔUθ

(iv)= 0

Answer 4 :

Since ΔHθ = ΔUθ + ΔngRT and ΔUθ = –X kJmol–1,

ΔHθ = (–X) + ΔngRT.

ΔHθ < ΔUθ

Therefore,alternative (iii) is correct.

Question 5 :

The enthalpy of combustion of methane,graphite and dihydrogen at 298 K are, –890.3 kJ mol–1 –393.5 kJ mol–1, and –285.8 kJ mol–1 respectively.Enthalpy of formation of CH4(g) will be

(i) –74.8 kJ mol–1            (ii) –52.27 kJ mol–1

(iii) +74.8 kJ mol–1          (iv) +52.26 kJ mol–1.

Answer 5 :

According to the question,

Thus, the desired equation is the one thatrepresents the formation of CH4 (g) i.e.,

Enthalpy of formation of CH4(g= –74.8 kJ mol–1

Hence,alternative (i) is correct.

Question 6 :

A reaction, A + B → C + D + q isfound to have a positive entropy change. The reaction will be

(i) possible at hightemperature

(ii) possible only at lowtemperature

(iii) not possible at anytemperature

(iv)possible at any temperature

Answer 6 :

For a reaction to be spontaneous, ΔG shouldbe negative.

ΔG = ΔH – TΔS

According to the question,for the given reaction,

ΔS = positive

ΔH = negative (since heat isevolved)

ΔG = negative

Therefore, the reaction isspontaneous at any temperature.

Hence,alternative (iv) is correct.

Question 7 :

Ina process, 701 J of heat is absorbed by a system and 394 J of work is done bythe system. What is the change in internal energy for the process?

Answer 7 :

According to the first lawof thermodynamics,

ΔU = q + W (i)

Where,

ΔU = change in internal energyfor a process

q = heat

W = work

Given,

q = + 701 J (Since heat is absorbed)

W = –394 J (Since work is done by thesystem)

Substituting the values inexpression (i), we get

ΔU = 701 J + (–394 J)

ΔU = 307 J

Hence,the change in internal energy for the given process is 307 J.

Question 8 :

The reaction of cyanamide, NH2CN(s),with dioxygen was carriedout in a bomb calorimeter, and Δwas found to be –742.7kJ mol–1at 298 K. Calculate enthalpy change for thereaction at 298 K.

Answer 8 :

Enthalpy change for a reaction (ΔH) is given by theexpression,

ΔH = ΔU + ΔngRT

Where,

ΔU = change in internalenergy

Δng = change in numberof moles

For the given reaction,

Δng = ∑ng (products) – ∑ng (reactants)

= (2 – 1.5) moles

Δng = 0.5 moles

And,

ΔU = –742.7 kJ mol–1

T = 298 K

R = 8.314 × 10–3 kJ mol–1 K–1

Substituting the values in the expression ofΔH:

ΔH = (–742.7 kJ mol–1) + (0.5 mol) (298 K)(8.314 × 10–3 kJ mol–1 K–1)

= –742.7 + 1.2

ΔH = –741.5 kJ mol–1

Question 9 :

Calculate the number of kJ of heat necessaryto raise the temperature of 60.0 g of aluminium from 35°C to 55°C. Molar heatcapacity of Al is 24 J mol–1 K–1.

Answer 9 :

From the expression of heat (q),

q = m. c. ΔT

Where,

c = molar heat capacity

m = mass of substance

ΔT = change in temperature

Substituting the values in the expressionof q:

q = 1066.7 J

q = 1.07 kJ

Question 10 :

Calculate the enthalpy change on freezing of1.0 mol of water at 10.0°C to ice at –10.0°C. Δfus= 6.03 kJ mol–1 at 0°C.

Cp[H2O(l)] = 75.3 J mol–1 K–1

Cp[H2O(s)] = 36.8 J mol–1 K–1

Answer 10 :

Total enthalpy changeinvolved in the transformation is the sum of the following changes:

(a) Energy change involvedin the transformation of 1 mol of water at 10°C to 1 mol of water at 0°C.

(b) Energy change involvedin the transformation of 1 mol of water at 0° to 1 mol of ice at 0°C.

(c) Energy change involvedin the transformation of 1 mol of ice at 0°C to 1 mol of ice at –10°C.

= (75.3 J mol–1 K–1) (0 – 10)K + (–6.03 × 103 J mol–1) + (36.8 J mol–1 K–1) (–10 – 0)K

= –753 J mol–1 – 6030 J mol–1 – 368 J mol–1

= –7151 J mol–1

= –7.151 kJ mol–1

Hence,the enthalpy change involved in the transformation is –7.151 kJ mol–1.


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