Laws Of Thermodynamics(11th And 12th > Physics ) Questions and answers for exam Preparation

11th And 12th > Physics

LAWS OF THERMODYNAMICS MCQs

Total Questions : 30 | Page 1 of 3 pages

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 2.

Tony now wants to design the engine with efficiency greater than that of Carnot's Engine, will he succeed? (Assuming his scientific work to be governed by the laws of physics that apply in the real world)

Yes, of course, he is Iron Man!
Yes, he has state of the art equipment at his disposal, Carnot was old.
No, it violates 1 $^{s t}$ law of thermodynamics.
No, the efficiency of Carnot's engine is the theoretical upper limit of efficiency, he can at best match it.

Discuss Question

Answer: Option D. -> No, the efficiency of Carnot's engine is the theoretical upper limit of efficiency, he can at best match it.
:
D
The efficiency of a Carnot's engine is the highest possible theoretical efficiency of any engine, any other engine can only match it, not improve upon it. Any engine with a better efficiency can be proved to violate the 2

^{n d}

law of thermodynamics.

Question 3.

Tony Stark wants to build an engine to propel his Iron Man suit; he wants to achieve an efficiency of 1.5 to make the suit lighter and faster! Which of the following laws of physics would he be violating?

1 $^{s t}$ law of Thermodynamics
2 $^{n d}$ law of Thermodynamics
Both (a) and (b)
Neither (a) and (b)

Discuss Question

Answer: Option C. -> Both (a) and (b)
:
C
Efficiency greater than 1 means that the work output of the engine is greater than the heat input! This violates the conservation of energy, or the 1

^{s t}

law of thermodynamics. Also the 2

^{n d}

law prohibits the efficiency to be equal to 1, hence it is also being violated.

Question 4.

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 5.

An ideal gas is taken through the process ABCDA as shown. Which process(s) is/are isobaric?

AB and CD
BC and DA
All processes
None of above

Discuss Question

Answer: Option B. -> BC and DA
:
B
Isobaric process is constant pressure process, on the graph AB and CD represent constant pressure lines.

Question 6.

Figure below represents a Carnot engine that works between temperatures $T_{1}$ 400 K and $T_{2}$ 150 K and drives aCarnot refrigerator that works between temperatures $T_{3}$ 325 K and $T_{4}$ 225 K.What is the ratio $\frac{Q 3}{Q 1}$ ?

$\frac{104}{85}$
$\frac{104}{95}$
$\frac{85}{104}$
$\frac{95}{104}$

Discuss Question

Answer: Option C. ->

\frac{85}{104}

:
C

Let us consider the efficiency of the heat engine first.

$η_{H . E} = 1 - \frac{T_{2}}{T_{1}}$

$= 1 - \frac{150}{400} = \frac{5}{8}$

$∴ W = η_{H . E} \times Q_{1}$

$= \frac{5}{8} Q_{1}$ .......................(1)

Now this work is utilized by the refrigerator to add heat to the hot body from the cold body. The efficiency of this refrigerator is calculated using.

$η_{r} = 1 - \frac{T_{3}}{T_{4}}$

$= 1 - \frac{225}{325}$

$= \frac{4}{13}$ ...........(2)

We know that $η_{r} = \frac{Q_{2}}{w}$

$\Rightarrow Q_{2} = w \times η_{r}$ ............(3)

From $1^{s t}$ law of thermodynamics we know

$Q_{3} = Q_{2} + w$

$= w + w \times η_{r}$ - - - - - - (4)

Substituting for w and $e t a_{r}$ in (4) we get

$Q_{3} = \frac{5}{8} Q_{1} + \frac{5}{8} Q_{1} \times \frac{4}{13}$

$= Q_{1} (\frac{65 + 20}{8 \times 13})$

$= Q_{1} \times \frac{85}{104}$

$\frac{Q_{3}}{Q_{1}} = \frac{85}{104}$

Question 7.

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 8.

What is the efficiency of a Carnot engine operating between higher temperature $T_{H}$ and lower temperature $T_{L}$ ?

$T_{H} - T_{L}$
$\frac{T_{H} - T_{L}}{T_{L}}$
$\frac{T_{H} - T_{L}}{T_{H}}$
$\frac{T_{L} - T_{H}}{T_{L}}$

Discuss Question

Answer: Option C. ->

\frac{T_{H} - T_{L}}{T_{H}}

:
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

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 9.

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 10.

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 = $n C_{V} Δ T$
Translational kinetic energy < rotational 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.

(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, $K_{T r} = \frac{3}{2} k T$ ,