Questions and answers for exam Preparation

MCQs

Laws Of Thermodynamics

Total Questions : 1462 | Page 1 of 147 pages

Question 1. Now Tony settles upon a Carnot engine. Calculate the efficiency of the Carnot engine if the temperature it operates between are 350 K and 250 K.

27
57
37
47
along the line CD

Discuss Question

Answer: Option A. -> 27
:
A
The efficiency of a Carnot engine is given by the formula

η_{H . E} = 1 - \frac{T_{L}}{T_{H}}

Substituting for known values we get

η_{H . E} = 1 - \frac{250}{350} = \frac{2}{7}

Question 2. An ideal gas in a heat engine executes the cycle shown. Where is the temperature of the gas maximum?

along the line BC
at point B
at point C
at point D
along the line CD

Discuss Question

Answer: Option C. -> at point C
:
C
Temperature is maximum at point C. for ideal gas, pV = NRT, so T is max, when pV is max, which corresponds to the point furthest from the origin.

Question 3. Again, the 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, what is the efficiency of this engine?

Discuss Question

Answer: Option C. -> 82%
:
C

Q_{H} = W + Q_{L} = 50000 J

e = \frac{E n e r g y o u t p u t}{E n e r g y i n p u t} = \frac{W}{Q_{H}} = \frac{41000 J}{50000 J} = 82 o r 82 %

Question 4. The figure here shows five paths traversed by a gas on a P-V diagram. Rank the paths according to the change in internal energy of the gas, greatest first.

5 > 4 = 3 > 2 = 1
5 > 4 = 3 > 1 = 2
4 = 3 > 5 > 1 = 2
5 > 4 = 3 = 2 = 1
along the line CD

Discuss Question

Answer: Option D. -> 5 > 4 = 3 = 2 = 1
:
D
For any ideal gas, the internal energy can be related to the temperature as -

E_{i n t} = n C_{v} T

\Rightarrow Δ E_{i n t} = n C_{v} Δ T

(for a change of state).
Thus the change in internal energy is solely dictated by the change in temperature.Since, all the processes 1, 2, 3, and 4 begin and end at the same temperatures, their

Δ T^{'} s (a n d, Δ E_{i n t} = n C_{v} (T_{2} - T_{1}))

are the same. Process 5, on the other hand goes through a larger

Δ T (T_{3} - T_{1})

, hence gains the highest internal energy.

Question 5. A typical dorm room or bedroom contains about 2500 moles of air. Find the change in the internal energy of this much air when it is cooled from

{35.0}^{\circ} C

{26.0}^{\circ} C

at a constant pressure of 1.00 atm. Treat the air as an ideal gas with

γ = 1.400

4.68 × 105 J
5.14 × 104 J
−4.68 × 105 J
−5.14 × 104 J

Discuss Question

Answer: Option C. -> −4.68 × 105 J
:
C
The change in internal energy

Δ E_{i n t} = n C_{V} Δ T

. We need to obtain

C_{V}

from the given information. Now,

γ = [\frac{C_{P}}{C_{V}}]

, and,

C_{P} - C_{V} = R

\Rightarrow γ = [\frac{C_{V} + R}{C_{V}}] = [1 + \frac{R}{C_{V}}]

\Rightarrow \frac{R}{C_{V}} = γ - 1

\Rightarrow C_{V} = [\frac{R}{γ - 1}]

Therefore,

Δ E_{i n t} = n C_{V} Δ T

= 2500 \times (\frac{8.314}{1.4 - 1}) \times (26 - 35) J

= - 4.68 \times 10^{5} J

.
It's a negative change, meaning the internal energy has decreased with cooling, which is consistent with theory. In fact, if an AC had to do this cooling, its job would be take

4.68 \times 10^{5} J

from the internal energy in the air and dump it outside. This will be dealt with in greater detail when we study the second law of thermodynamics.

Question 6. 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 7. 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}

Figure Below Represents A Carnot Engine That Works Between T...

10485
10495
85104
95104

Discuss Question

Answer: Option C. -> 85104
:
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 8. 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}

10485
10495
85104
95104

Discuss Question

Answer: Option C. -> 85104
:
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 9. A window air conditioner is brought into the center of a dorm room and turned on. After several minutes, the temperature of the room:

increases
decreases
remains constant
depends on the brand of the AC

Discuss Question

Answer: Option A. -> increases
:
A
Air conditioners only work if you put them in the window, so they can eject heat from the room into the great outdoors. Any power input to the air conditioner (from you electrical supply) is turned into heat that is output, thereby increasing the room temperature. Mathematically, the heat rejected into the room,

Q_{1} = W + Q_{2}

, where

Q_{2}

is the heat gained from the refrigerator and W the work input. Similarly, ou can't cool your kitchen by opening the fridge door.

Question 10. A heat engine executes one complete cycle. The change in entropy of the engine (just the engine, not the external thermal reservoirs) is:

zero
greater than zero
less than zero
depends on the type of engine
along the line CD

Discuss Question

Answer: Option A. -> zero
:
A
In one complete cycle, the engine returns to its original state. So all macroscopic variables (N, V, T, S etc) must return to their original state and the change is zero. Remember, even though heat is a path variable, the entropy S is a state variable, hence this stamen will be true even for an irreversible process.