Questions and answers for exam Preparation

MCQs

Laws Of Thermodynamics

Total Questions : 1462 | Page 3 of 147 pages

Question 21. What is the efficiency of a Carnot engine operating between higher temperature

T_{H}

and lower temperature

T_{L}

TH−TL
TH−TLTL
TH−TLTH
TL−THTL

Discuss Question

Answer: Option C. -> TH−TLTH
:
C
The work done by the gas over the entire cycle divided by the heat consumed from the hot reservoir is the efficiency of the Carnot engine.
Let us consider each isothermal process and find the work done in each of them.
Process a - b (isothermal)
Work done in an isothermal process is given by the formula

W = n R T ℓ_{n} (\frac{v_{2}}{v_{1}})

In this case

W_{1} = n R T ℓ_{n} (\frac{v_{n}}{v_{b}})

. . . . (1)
Similarly the work done in process c - d is

W_{3} = n R T ℓ_{n} (\frac{v_{c}}{v_{d}})

. . . . (2)
Since both of these process are isothermal

Q_{L} = W_{3} . . . (4) a n d Q_{H} = W_{1} . . .

(5)
The Carnot engine is represented below

What Is The Efficiency Of A Carnot Engine Operating Between ...

Applying law of conservation of energy we get

Q_{H} - Q_{L} = W

Dividing throughout by

Q_{H}

we get

1 - \frac{Q_{L}}{Q_{H}} = \frac{W}{Q_{H}} = η

. . . . .(6)
Substituting for

Q_{H}

and

Q_{L}

in(6)
We get

1 - \frac{n R T_{L} l_{n} (\frac{v_{c}}{v_{d}})}{n R T_{H} l_{n} (\frac{v_{b}}{v_{a}})} = η

\Rightarrow η = \frac{1 - \frac{T_{L}}{T_{H}} l_{n} (\frac{v_{c}}{v_{d}})}{(\frac{v_{b}}{v_{a}})}

In process b - c since it is adiabatic the state variables T and V are connected using the relation

T_{b}^{(\frac{1}{γ - 1})} \times V_{b} = T_{c}^{(\frac{1}{γ - 1})} \times V_{c}

......(8)
and similarly

T_{d}^{(\frac{1}{γ - 1})} \times V_{d} = T_{a}^{(\frac{1}{γ - 1})} \times V_{a}

.......(9)
From (8) and we get

\frac{v_{c}}{v_{d}} = \frac{v_{b}}{v_{a}}

\Rightarrow l_{n} (\frac{v_{c}}{v_{d}}) l_{n} (\frac{v_{b}}{v_{a}})

.......(10)
Substituting (10) in (9) we get

e t a = 1 - T_{L}

η = 1 - \frac{T_{L}}{T_{H}}

Question 22. An engine does 41,000 J of useful work. If it sinks 9,000 J of energy in the form of heat to its cold reservoir, find the heat energy supplied to the engine's hot reservoir?

41000 J
9000 J
32000 J
50000 J

Discuss Question

Answer: Option D. -> 50000 J
:
D

Q_{H O T} = W + Q_{C O L D} = 41000 J + 9000 J = 50000 J

Question 23. Imagine a Carnot engine that operates between the temperatures

T_{H}

850 K and

T_{L}

300 K. The engine performs 1200 J of work each cycle, which takes 0.25 s. What is the efficiency of this engine?

1119
1117
819
617

Discuss Question

Answer: Option B. -> 1117
:
B
The efficiency of the engine can be calculated by the formula in the hint

η = 1 - \frac{300}{850}

= 1 - \frac{6}{17} = \frac{11}{17}

Question 24. Along the line AB :

heat was input
heat was extracted
no heat was transferred
Can't be said

Discuss Question

Answer: Option A. -> heat was input
:
A
Along AB heat was input. Along the line AB, the quantity pV increased, so NkT increased so

Δ U = N k Δ T

was positive.

Δ U = Q + W

, and W = 0 since

p Δ V = 0

. So

Δ U = Q

is positive.

Question 25. In the P-Vdiagram show, moles of a gas are taken through a process from state i to f, at constant pressure. Part of the heat must have gone into doing the work

P Δ V

, and the rest in increasing the internal energy. If the molar specific heat at constant volume for the gas is

C_{V}

, and the initial total internal energy is

U_{i}

, what is the final internal energy

U_{f}

In The P-Vdiagram Show, Moles Of A Gas Are Taken Through A P...

Uf = Ui(since total internal energy is always conserved)
Uf = Ui + nCvT
Uf = Ui + nCvΔT
Uf = Ui + CvΔT

Discuss Question

Answer: Option D. -> Uf = Ui + CvΔT
:
D
: In the diagram, the two lines represent isotherms for temperatures T and

(T + Δ T)

- any point on either line will be at the same temperature. If a process takes the state of a gas from a point on one isotherm to a point on another isotherm, the change in internal energy will be related to the temperature difference only, and nothing else, given by the relation -

Δ E_{i n t} = n C_{v} Δ T

Therefore in our case,

U_{f} = U_{i} + Δ E_{i n t}

\Rightarrow U_{f} = U_{i} + n C_{v} Δ T

Question 26. A refrigerator uses 400 J of work to remove 200 J of heat from its contents. How much heat must it reject to its surroundings?

400 J
200 J
600 J
800 J

Discuss Question

Answer: Option C. -> 600 J
:
C
The refrigerator must reject an amount of heat to the room equal to the sum of the work done by the refrigerator and the heat removed from its contents or 400 J + 200 J = 600 J

Question 27. In one cycle of the engine, the net change in the entropy of both reservoirs (hot and cold together) is:

zero
greater than or equal to zero
less than zero
depends on the type of engine

Discuss Question

Answer: Option B. -> greater than or equal to zero
:
B
Greater than or equal to zero. By the

2^{n d}

Law, the entropy of the universe (engine + hot bath + cold bath) can never decrease. The entropy of the engine remains constant (see previous question) so the entropy of the reservoirs must increase. In the case of an ideal Carnot cycle, the process is reversible and the net entropy change is zero.

Question 28. We know that, for an ideal monoatomic gas, the total internal energy is a result of the translational kinetic energy only. For the oxygen molecule

O_{2}

(assume it doesn't oscillate along the bond axis), which rotates as well as translates, choose the correct statement(s) from the following

Change in translational kinetic energy of a sample of n moles = nCVΔT
Translational kinetic energy
Translational kinetic energy > rotational kinetic energy
Translational kinetic energy = rotational kinetic energy.

Discuss Question

Answer: Option C. -> Translational kinetic energy > rotational kinetic energy
:
C
The equipartition theorem postulates - each degree of freedom contributes an energy equal to

(\frac{1}{2}) k T

to the total internal energy of a molecule, depending on the temperature T. Let's count the number of degrees of freedom for an

O_{2}

molecule.

We Know That, For An Ideal Monoatomic Gas, The Total Interna...

(1) Possible directions of translational motion are the three axes, x, y and z. Thus, number of translational degrees of motion = 3.
(2) Given the negligible size of the atoms, rotation about the bond-axis will not be significant. It can only rotate about the two axes shown (perpendicular to the bond-axis). Thus number of rotational degrees of motion = 2.
Therefore, according the equipartition theorem,
(i) Translational Kinetic energy,