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In which of the paths between initial state i and final state f in the below figure is the work done on the gas the greatest?
Solution:-
The correct option is (D).
Work is a path function. So work done on the gas depends upon the path. The area under pv diagram gives the work done on the gas between initial state i and final state f. As the area under the path D is greater than the other path, therefore the paths between initial state i and final state f where the work done on the gas is greatest would be D. From the above observation we conclude that, option D is correct.
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A 1-kg block of ice at 0ºC is placed into a perfectly insulated, sealed container that has 2 kg of water also at 0ºC. The water and ice completely fill the container, but the container is flexible. After some time one can except that
(A) the water will freeze so that the mass of the ice will increase.
(B) the ice will melt so that the mass of the ice will decrease.
(C) both the amount of water and the amount of ice will remain constant.
(D) both the amount of water and the amount of ice will decrease.
The correct option is (B) the ice will melt so that the mass of the ice will decrease.
In accordance to second law of thermodynamics, if an irreversible process occurs in a closed system, the entropy of that system always increases; it never decreases. Entropy measures the state of disorder. As the ice is more order than water, therefore the ice will melt so that the mass of the ice will decrease. From the above observation we conclude that, option (B) is correct.
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Which of the following processes must violate the first law of thermodynamics? (There may be more than one answer!)
(A) W > 0, Q < 0, and ?E_{int} = 0
(B) W > 0, Q < 0, and ?E_{int} > 0
(C) W > 0, Q < 0, and ?E_{int} < 0
(D) W < 0, Q > 0, and ?E_{int} < 0
(E) W > 0, Q > 0, and ?E_{int} < 0
The correct option is (E).
In accordance to first law of thermodynamics, for a thermodynamic system, in which internal energy is the only type of energy the system may have, the law of conservation of energy may be expressed as
Q + W = ΔE_{int}
Here Q is the energy transferred (as heat) between the system and environment, W is the work done on (or by) the system and ΔE_{int} is the change in the internal energy of the system.
By convention we have chosen Q to be positive when heat is transferred into the system and W to be positive when work is done on the system.
In the option (E), W > 0 and Q > 0. If we put this condition in the left hand side of the equation Q + W = ΔE_{int}, then the right hand side of the equation must be greater than zero (ΔE_{int} > 0). Thus W > 0, Q > 0, and ΔE_{int} < 0 must violate the first law of thermodynamics. Therefore option (E) is correct.
A spherical constant temperature heat source of radius r_{1} is at the center of a uniform solid sphere of radius r_{2}. The rate at which heat is transferred through the surface of the sphere is propertional to
(A) r_{2}^{2} – r_{1}^{2}
(B) r_{2} – r_{1}
(C) ln r_{1} – ln r_{2}
(D) 1/r_{2} – 1/r_{1}
(E) (1/r_{2} – 1/r_{1})^{-1}
The correct option is (A).
The rate H at which heat is transferred through the slab is,
(a) directly proportional to the area (A) available.
(b) inversely proportional to the thickness of the slab Δx.
(c) directly proportional to the temperature difference ΔT.
So, H = kA ΔT/ Δx
Where k is the proportionality constant and is called thermal conductivity of the material.
From above we know that, the rate H at which heat is transferred through the slab is directly proportional to the area (A) available.
Area A of solid sphere is defined as,
A = 4πr^{2}
Here r is the radius of sphere.
So, the area A_{1} of uniform small solid sphere having radius r_{1} will be,
A_{1} = 4πr_{1}^{2}
And, the area A_{2} of uniform large solid sphere having radius r_{2} will be,
A_{2} = 4πr_{2}^{2}
Thus the area A from which heat is transferred through the surface of the sphere will be the difference of area of uniform large solid sphere A_{2} and small solid sphere A_{1}.
So, A = A_{2} - A_{1 }= 4πr_{2}^{2} - 4πr_{1}^{2 }= 4π (r_{2}^{2} - r_{1}^{2})
Since the rate H at which heat is transferred through the slab is directly proportional to the area (A) available, therefore the rate at which heat is transferred through the surface of the sphere is proportional to r_{2}^{2} - r_{1}^{2}.
From the above observation we conclude that, option (A) is correct.
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What would be the most likely value for C_{T}, the molar heat capacity at constant temperature?
(A) 0
(B) 0 < C_{T} < C_{v}
(C) C_{v} < C_{T} < C_{p}
(D) C_{T} =
Molar heat capacity at constant temperature (C_{T}) is defined as the amount of heat required to raise the temperature of 1 g of gas through 1° C keeping its temperature constant. But it is impossible to rise the temperature of 1 g of gas through 1° C keeping its temperature constant. So the two statements contradict to each other. Thus the most likely value for molar heat capacity at constant temperature (C_{T}) will be zero. Therefore option (A) is correct.
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Which type of ideal gas will have the largest value for C_{p} – C_{v}?
(A) Monoatomic
(B) Diatomic
(C) Polyatomic
(D) The value will be the same for all.
The specific heat at constant volume (C_{v}) for monoatomic gas is,
C_{v} = 3/2 R
The specific heat at constant volume (C_{v}) for diatomic gas is,
C_{v} = 5/2 R
The specific heat at constant volume (C_{v}) for polyatomic gas is,
C_{v} = 3 R
But we know that, for an ideal gas,
C_{p} - C_{v} = R
So,the specific heat at constant pressure (C_{p}) for monoatomic gas will be,
C_{p} = C_{v} + R = 3/2 R + R = 5/2 R
the specific heat at constant pressure (C_{p}) for diatomic gas will be,
C_{p} = C_{v} + R = 5/2 R + R = 7/2 R
the specific heat at constant pressure (C_{p}) for polyatomic gas will be,
C_{p} = C_{v} + R = 3 R + R = 4 R
Therefore,
C_{p} - C_{v} for monoatomic gas will be,
C_{p} - C_{v} = 5/2 R - 3/2 R = R
C_{p} - C_{v} fo diatomic gas will be,
C_{p} - C_{v} = 7/2 R - 5/2 R = R
C_{p} - C_{v} fo polyatomic gas will be,
C_{p} - C_{v} = 4 R - 3 R = R
From the above observation we conclude that, the value of C_{p} - C_{v} will be same for all ideal gas. Therefore option (D) is correct.
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Consider the following processes that can be done on an ideal gas: constant volume, ΔV = 0; constant pressure, Δp = 0; and constant temperature, ΔT = 0.
(a) For which process does W = 0?
(b) For which process does Q = 0?
(c) For which of these process does W + Q = 0?
(d) For which of these process does ΔE_{int} = Q?
(e) For which of these process does ΔE_{int} = W?
(A) ΔV = 0 (B) Δp = 0 (C) ΔT = 0 (D) None of these
(a) The correct option is (A) ΔV = 0.
The change in work done ΔW is defined as,
ΔW = pΔV
Therefore W = 0 for the process where ΔV = 0. From the above observation we conclude that, option (A) is correct.
In adiabatic process the exchange of heat with surrounding is zero (Q=0). In this process, the change of volume, pressure and temperature occurs. Therefore option (D) is correct.
For a thermodynamic system, in which internal energy is the only type of energy the system may have, the law of conservation of energy may be expressed as,
Here Q is the energy transferred (as heat) between the system and its environment, W is the work done on or by the system and ΔE_{int} is the change in the internal energy of the system.
As W + Q = 0, it signifies that ΔE_{int} = 0 and it is possible only when ΔT = 0. Therefor option (C) is correct.
As ΔE_{int} = Q, it signifies that W=0.
As ΔE_{int} = W, it signifies that Q=0.
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Consider an ideal heat pump and a perfect electric heater. The electric heater converts 100% of the electrical energy into heat energy; the heat pump converts 100% of the electrical energy into work, which then powers a Carnot refrigerator. Which is the more “efficient” way to heat a home? (Ignore maintenance or start-up costs.)
(A) The electric heater is always more efficient.
(B) The heat pump is always more efficient.
(C) The heat pump is more efficient if the outside temperature is not too cold.
(D) The heat pump is more efficient if the outside temperature is not too cold.
The correct option is (C).
A heat pump is a device that acting as a Carnot refrigerator can heat a house by transferring heat energy from the outside to the inside of the house; the process is driven by work done on the device. So, heat pump, which is also a refrigerator, is an air conditioner that can be operated in reverse to heat room. By heating the room using a perfect electric heater, there is some loss of thermal energy due to heating the coil of the electric heater.
The value of coefficient of performance K becomes larger as the temperature of the two reservoirs becomes more nearly the same. Thus heat pumps are more efficient in temperate climates than in climates where the outside temperature fluctuates between wide limits. From the above observation we conclude that, the heat pump is more efficient if the outside temperature is not too warm. Thus option (C) is correct.
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A real engine operates at 75% of the efficiency of a Carnot engine operating betwen the same two temperatures. This engine has a power output of 100 W and discharges heat into the 27ºC low-temperature reservior at a rate of 300 J/s. what is the temperature of the high-temperature reservior?
(A) 27ºC (B) 77ºC (C) 127ºC (D) 177ºC
The correct option is (C) 127^{ °} C.
The efficiency ε of a Carnot engine is defined as,
ε = |Q_{H}| -|Q_{L}|/|Q_{H}|
=1- |Q_{L}|/|Q_{H}|
= 1- T_{L}/T_{H}
Here Q_{L} is the heat at lower temperature, Q_{H} is the heat at higher temperature, T_{L} is the lower temperature and T_{H} is the higher temperature.
As the real engine has a power output of 100 W (100 J/s) and discharge heat into the 27^{ °} C low-temperature reservoir at a rate of 300 J/s.
Here, |Q_{H}| -|Q_{L}| = 100 J
And
|Q_{L}| = 300 J
So, |Q_{H}| -300 J = 100 J
|Q_{H}| = 400 J
To obtain the efficiency ε of this real engine, substitute 300 J for Q_{L} and 400 J for Q_{H} in the equation ε = 1- |Q_{L}/Q_{H}|,
ε = 1- |Q_{L}/Q_{H}|
= 1-300 J/400 J
= 1/4
To obtain the temperature of the high-temperature region T_{H}, substitute 1/4 for ε, 27^{ °} C for T_{L} in the equation ε =1- T_{L}/T_{H} we get,
1/4=1- T_{L}/T_{H}
= 1-27^{ °} C/ T_{H}
= 1- (27+273) K/ T_{H}
= 1- 300 K/ T_{H}
So,300 K/ T_{H} = ¾
T_{H }= 400 K
= (400 -273)^{°} C
= 127 ^{°} C
Therefore the temperature of the high-temperature region T_{H} would be 127 ^{°} C. From the above observation we conclude that, option (C) is correct.
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A real has an efficiency of 33%. The engine has a work output of 24 J per cycle.
(a) How much heat energy is extracted from the high temperature reservior per cycle?
(A) 8 J (B) 16 J (C) 48 J (D) 72 J (E) The question can be answered only if the engine is a carnot engine
(b) How much heat energy is discharged into the low-temperature reservior per cycle?
(c) For this engine T_{L }= 27ºC, What can be concluded about T_{H}?
(A) T_{H} = 450ºC (B) T_{H} = 177ºC
(C) T_{H} > 177ºC (D) T_{H} < 177ºC
(E) 177ºC < T_{H} < 450ºC
(a) The correct option is (D) 72 J.
As, ε = |Q_{H}| -|Q_{L}|/|Q_{H}|
= ΔW/|Q_{H}| (Since, |Q_{H}| -|Q_{L}| = ΔW)
To obtain the amount of heat energy is extracted from the higher temperature |Q_{H}|, substitute 33% for ε and 24 J for ΔW in the equation ε = ΔW/|Q_{H}|, we get,
ε = ΔW/|Q_{H}|
33/100 = 24 J/|Q_{H}|
Or, |Q_{H}| = (2400/33) J
= 72.73 J
Rounding off to two significant figures, the amount of heat energy is extracted from the higher temperature |Q_{H}| would be 72 J. From the above observation we conclude that, option (D) is correct.
To obtain the heat energy is discharged into the low-temperature reservoir per cycle |Q_{L}|, substitute 72 J for |Q_{H}| and 33% for ε in the equation ε =1- |Q_{L}|/|Q_{H}|,
ε =1- |Q_{L}|/|Q_{H}|
33/100 = 1-|Q_{L}|/72 J
|Q_{L}|/72 J = 1-33/100
= 67/100
So,|Q_{L}| = (67/100)×72 J
= 48.24 J
Rounding off to two significant figures, the heat energy is discharged into the low-temperature reservoir per cycle |Q_{L}| would be 48 J. From the above observation we conclude that, option (C) is correct.
To obtain T_{H}, substitute 33% for ε, 27^{ °} C for T_{L} in the equation ε =1- T_{L}/T_{H} we get,
33/100 =1- T_{L}/T_{H}
27^{ °} C/ T_{H} =1-33/100
=67/100
T_{H} =27^{ °} C×100/67
=40.30^{ °} C
<177^{ °} C
Therefore the value of T_{H} will be <177^{ °} C. From the above observation we conclude that, option (D) is correct.
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A Carnot engine discharges 3 J of heat into the low temperature reservior for every 2 J of work output.
(a) What is the efficiency of this Carnot engine?
(A) 1/3 (B) 2/5 (C) 3/5 (D) 2/3
(b) For this engine T_{L} = 27ºC, What can be concluded about T_{H}?
(A) T_{H} = 627ºC (B) T_{H} = 227ºC
(C) T_{H} > 627ºC (D) T_{H} < 227ºC
(E) 227ºC < T_{H} < 627ºC
(a) The correct option is (A) 1/3.
To obtain the efficiency ε of this Carnot engine, substitute 2 J for Q_{L} and 3 J for Q_{H} in the equation ε = 1- |Q_{L}/Q_{H}|,
= 1-2 J/3 J
= 1/3
Thus the efficiency of this Carnot engine would be 1/3. From the above observation we conclude that, option (A) is correct.
Thus the efficiency of this Carnot engine would be 1/3.
To obtain T_{H}, substitute 1/3 for ε, 27^{ °} C for T_{L} in the equation ε =1- T_{L}/T_{H} we get,
1/3=1- T_{L}/T_{H}
T_{H} = 27^{ °} C×3/2
=40.5^{°} C
<227^{ °} C
Therefore the value of T_{H} will be <227^{ °} C. From the above observation we conclude that, option (D) is correct.
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Which of the following is a consequence of the second law of thermodynamics?
(A) Heat can flow only from high temperature to low temperature.
(B) Objects in contact will tend toweard having the same temperature.
(C) Any system that produces order from disorder must have
In accordance to second law of thermodynamics, when changes occur due to external influence within a closed system its entropy either increases (for irreversible process) or remains constant (for reversible process). It never decreases. Entropy is often associated with disorder and the second law of thermodynamics is sometimes cast as a statement that the disorder of a closed system always increases. The final state, with the randomly directed motions of its molecules, is more disorder than the initial state, with the directed motions of a substantially large number of its molecules. Thus any system that produces order from disorder must have an external influence. Therefore option (C) is correct.
A block of aluminum originally at 80º C is placed into an insulated container of water originally at 25ºC. After a while the system reaches an equilibrium temperature of 31ºC.
(a) During the process
(A) ΔS_{aluminum} > 0 (B) ΔS_{aluminum} (C) ΔS_{aluminum} < 0
(b) During this process
(A) ΔS_{water} > 0 (B) ΔS_{water} = 0 (C) ΔS_{water} < 0
(c) During this process
(A) |ΔS_{water}| > |ΔS_{aluminum}|
(B) |ΔS_{water}| = |ΔS_{aluminum}|
(C) |ΔS_{water}| < |ΔS_{aluminum}|
(a) The correct option is (C) ΔS_{aluminum}<0.
Entropy change (ΔS) for a reversible process is defined as,
ΔS = ΔQ/T
Where, ΔQ is the change in heat energy that is transferred into or out of the closed system at constant temperature T.
Whenever an irreversible process takes place, certain amount of energy which could have been utilized for doing useful work changes to a form in which it becomes unavailable. Entropy is a measure of disorder. Thus with disorder, entropy increases. The initial temperature of the aluminum block is 80^{°}C. But after some time the system reaches an equilibrium temperature of 31^{°}C. Therefore the aluminum block will colder than its original state. This signifies that, the change in entropy of the block of aluminum will be less than zero (ΔS_{aluminum}<0). From the above observation we conclude that option (C) is correct.
(b) The correct option is (A) ΔS_{water}>0.
Whenever an irreversible process takes place, certain amount of energy which could have been utilized for doing useful work changes to a form in which it becomes unavailable. Entropy is a measure of disorder. Thus with disorder, entropy increases. The initial temperature of water is 25° C. But after some time the system reaches an equilibrium temperature of 31^{°}C. Therefore the water will be much hotter than its original state. This signifies that, the change in entropy of the water will be greater than zero (ΔS_{water}>0). From the above observation we conclude that option (A) is correct.
(c) The correct option is (A) |ΔS_{water}| > | ΔS_{aluminum}|
In accordance to second law of thermodynamics, when changes occur within a closed system its entropy increases for an irreversible process. To satisfy the second law of thermodynamics in the system, the net change in entropy of water must be greater than the net change in entropy of aluminum (|ΔS_{water}| > | ΔS_{aluminum}|). From the above observation we conclude that, option (A) is correct.
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For which of the following process is the entropy change zero?
(A) Isobaric (B) Isothermal
(C) Adiabatic (D) Constant volume
(E) None of these, since ΔS > 0 for all process.
The correct option is (C) adiabatic.
From equation ΔS = ΔQ/T, the entropy change will be zero when
ΔQ/T = 0
If the above condition is satisfied, ΔQ =0 as T is constant.
This shows that the change in the entropy is constant.
So, the process will be adiabatic process as the amount of heat remains constant. Thus in an adiabatic process the entropy change will be zero. Therefore option (C) is correct.
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Ten identical particles are to be divided up into two containers.
(a) How many microstates belong to the configuration of three particles in one container and seven in the other?
(A) 120 (B) 30240 (C) 3628800 (D) 6.310^{9}
(b) How many different configuration are possible?
(A) 1 (B) 11 (C) 120 (D) 1024 (E) 3628800
(c) What is the total number of microstates for the ten-particle system?
(d) Which configuration has the largest number of microstates?
(A) 0, 10 (B) 3, 7 (C) 4,6 (D) 5, 5
(a) The correct option is (A) 120.
The number of microstates that lead to a given configuration the multiplicity w of that configuration is defined as,
w = N!/N_{1}! N_{2}!
Here N = 10, N_{1} = 3 and N_{2} = 7.
So, w = N!/N_{1}! N_{2}!
= 10!/3! 7!
= 120
From the above observation we conclude that, the microstates belong to the configuration of three particles in one container and seven in other will be 120. Therefore option (A) is correct.
Ten identical particles are to be divided up into two containers. The total number of possible different configuration will be equal to N!.
Here N is the total number of particles.
As N = 10, N! = 10! = 3628800
From the above observation we conclude that, 3628800 different configuration are possible. Therefore option (E) is correct.
Ten identical particles are to be divided up into two containers. The total number of microstates for the ten particle system will be equal to 2^{N}.
As N = 10, 2^{N} = 2^{10} = 1024
From the above observation we conclude that, the total number of microstates for the ten particle system will be 1024. Therefore option (D) is correct.
In the configuration (0,10) , N = 10, N_{1} = 0 and N_{2} = 10.
So, w = N!/N_{1}! N_{2}! = 10!/0! 10! = 1
In the configuration (3,7) , N = 10, N_{1} = 3 and N_{2} = 7.
So, w = N!/N_{1}! N_{2}! = 10!/3! 7! = 120
In the configuration (4,6) , N = 10, N_{1} = 4 and N_{2} = 6.
So, w = N!/N_{1}! N_{2}! = 10!/4! 6! = 210
In the configuration (5,5) , N = 10, N_{1} = 5 and N_{2} = 5.
So, w = N!/N_{1}! N_{2}! = 10!/5! 5! = 252
From the above observation we conclude that, the configuration which has the largest number of microstates will be (5,5). Therefore option (D) is correct.
(A) Q_{1} = Q_{2}
(B) W_{1} + Q_{1} = Q_{2} + W_{2}
(C) W_{1} = W_{2}
(D) Q_{1} - W_{1} = Q_{2} - W_{2}
When the gas expands under adiabatic condition, its temperature:
(a) increases
(b) decreases
(c) does not change
(d) none of the above
In a reversible adiabatic change ΔS is
(A) infinity
(B) zero
(C) equal to C_{v}dT
(D) equal to nRln V_{2}/V_{1}
If the temperature of source increases, the efficiency of a Carnot’s engine:
(A) decreases
(B) increases
(C) remains the same
(D) may increase or decrease
If the temperature of the sink is decreased, the efficiency of a Carnot engine:
(D) first increase and then decrease
Heat cannot flow by itself from a colder body to hotter body is a statement of:
(A) 2^{nd} law of thermodynamics
(B) conservation of momentum
(C) conservation of mass
(d) first law of thermodynamics
D
B
A
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