26: Laws Of Thermodynamics : Cbse-xi-physics

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CBSE-XI-Physics

26: Laws of Thermodynamics

with Solutions - page 2

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Qstn #14
What should be the condition for the efficiency of a Carnot engine to be equal to 1?
Ans : Efficiency of Carnot engine, `` \eta =\frac{W}{{Q}_{1}}=\frac{{Q}_{1}-{Q}_{2}}{{Q}_{1}}``
Here,
W = Work done by the engine
Q₁ = Heat abstracted from the source
Q₂ = Heat transferred to the sink
Thus, for `` \eta =1,W={Q}_{1}or{Q}_{2}=0``, the total heat which is abstracted from the source gets converted into work.
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Qstn #15
When an object cools down, heat is withdrawn from it. Does the entropy of the object decrease in this process? If yes, is it a violation of the second law of thermodynamics stated in terms of increase in entropy?
Ans : When an object cools down, heat is withdrawn from it. Hence, the entropy of the object decreases. But the decrease in entropy leads to the transfer of energy to the surrounding. The second law is not violated here, which states that entropy of the universe always increases as the net entropy increases.
Here,
Net entropy = Entropy of object + Entropy of surrounding
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Section : ii

Qstn #1
The first law of thermodynamics is a statement of
(a) conservation of heat
(b) conservation of work
(c) conservation of momentum
(d) conservation of energy
digAnsr: d
Ans : (d) conservation of energy
Heat is a form of energy. Since the first law of thermodynamics deals with the conservation of heat, it actually refers to the conservation of energy in the broader sense.
The first law of thermodynamics is just the restatement of the law of conservation of energy. We observe that the energy supplied to a system will contribute to change in its internal energy and the amount of work done by the system on its surroundings.
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Qstn #2
If heat is supplied to an ideal gas in an isothermal process,
(a) the internal energy of the gas will increase
(b) the gas will do positive work
(c) the gas will do negative work
(d) the said process is not possible.
digAnsr: b
Ans : (b) the gas will do positive work
Internal energy of the gas, `` U=\int {C}_{v}dT``
`` ``
Here, Cv is the specific heat at constant volume and dT is the change in temperature.
If the process is isothermal, i.e. dT = 0, dU = 0.
Using the first law of thermodynamics, we get
`` \Delta Q=\Delta U+\Delta W``
`` But\Delta U=0``
`` \Rightarrow \Delta Q=+\Delta W``
Here, `` \Delta ``W is the positive work done by the gas.
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Qstn #3
Figure shows two processes A and B on a system. Let ▵Q₁ and ▵Q₂ be the heat given to the system in processes A and B respectively. Then
(a) ▵Q₁ > ▵Q₂
(b) ▵Q₁ = ▵Q₂
(c) ▵Q₁ < ▵Q₂
(d) ▵Q₁ ≤ ▵Q₂.
Figure
digAnsr: a
Ans :
(a) ∆Q₁ > ∆Q₂
Both the processes A and B have common initial and final points. So, change in internal energy, ∆U is same in both the cases. Internal energy is a state function that does not depend on the path followed.
In the P-V diagram, the area under the curve represents the work done on the system, ∆W. Since area under curve A > area under curve B, ∆W₁> ∆W₂.
Now,
`` \Delta {Q}_{1}=\Delta U+\Delta {W}_{1}``
`` \Delta {Q}_{2}=\Delta U+\Delta {W}_{2}``
`` \,\mathrm{\,But\,}\Delta {W}_{1}>\Delta {W}_{2}``
`` \Rightarrow \Delta {Q}_{1}>\Delta {Q}_{2}``
Here, ∆Q₁ and ∆Q₂ denote the heat given to the system in processes A and B, respectively.
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Qstn #4
Refer to figure. Let ▵U₁ and ▵U₂ be the changes in internal energy of the system in the process A and B. Then
(a) ▵U₁ > ▵U₂
(b) ▵U₁ = ▵U₂
(c) ▵U₁ < ▵U₂
(d) ▵U₁ ≠ ▵U₂.
digAnsr: b
Ans : (b) ∆U₁ = ∆U₂
The internal energy of the system is a state function, i.e. it only depends on the initial and final point of the process and doesn't depend on the path followed. Both processes A and B have common initial and final points. Therefore, change in internal energy in process A is equal to the change in internal energy in process B. Thus,
∆U₁ = ∆U₂= 0
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Qstn #5
Consider the process on a system shown in figure. During the process, the work done by the system
(a) continuously increases
(b) continuously decreases
(c) first increases then decreases
(d) first decreases then increases.
Figure
digAnsr: a
Ans :
(a) continuously increases
Work done by a system, `` W=\int PdV``
Here,
P = Pressure on the system
dV = change in volume.
Since dV is positive, i.e. the volume is continuously increasing, work done by the system also continuously increases.
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Qstn #6
Consider the following two statements.
(A) If heat is added to a system, its temperature must increase.
(B) If positive work is done by a system in a thermodynamic process, its volume must increase.
(a) Both A and B are correct.
(b) A is correct but B is wrong.
(c) B is correct but A is wrong.
(d) Both A and B are wrong.
digAnsr: c
Ans : (c) B is correct but A is wrong.
If heat is added to a system in an isothermal process, then there'll be no change in the temperature.
`` \,\mathrm{\,Work\,}\,\mathrm{\,done\,}\,\mathrm{\,by\,}\,\mathrm{\,a\,}\,\mathrm{\,system\,},∆W=P∆V``
`` \Rightarrow ∆W=\,\mathrm{\,positive\,}\Rightarrow ∆V=\,\mathrm{\,positive\,}``
Here,
P = Pressure
`` ∆``V = change in volume
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Qstn #7
An ideal gas goes from the state i to the state f as shown in figure. The work done by the gas during the process
(a) is positive
(b) is negative
(c) is zero
(d) cannot be obtained from this information.
Figure
digAnsr: c
Ans :
(c) is zero
Work done by the gas during the process, `` ∆W=P∆V``
`` ``
Here,
P = Pressure
`` ∆``V = change in volume.
Since the process described in the figure is isochoric, P = kT. As volume remains constant (`` ∆``V = 0), `` ∆``W = 0.
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Qstn #8
Consider two processes on a system as shown in figure.
Figure
The volumes in the initial states are the same in the two processes and the volumes in the final states are also the same. Let ▵W₁ and ▵W₂ be the work done by the system in the processes A and B respectively.
(a) ▵W₁ > ▵W₂.
(b) ▵W₁ = ▵W₂.
(c) ▵W₁ < ▵W₂.
(d) Nothing can be said about the relation between ▵W₁ and ▵W₂.
digAnsr: c
Ans :
(c) ∆W₁ < ∆W₂
Work done by the system, `` ∆W=P∆V``
Here,
P = Pressure in the process
`` ∆V`` = Change in volume during the process
Let V_i and V_f be the volumes in the initial states and final states for processes A and B, respectively. Then,
`` \Delta {W}_{1}={P}_{1}\Delta {V}_{1}``
`` \Delta {W}_{2}={P}_{2}\Delta {V}_{2}``
`` \,\mathrm{\,But\,}\Delta {V}_{2}=\Delta {V}_{1},\left[\left({V}_{{f}_{1}}-{V}_{{i}_{1}}\right)=\left({V}_{{f}_{2}}-{V}_{{i}_{2}}\right)\right]``
`` \Rightarrow \frac{\Delta {W}_{1}}{\Delta {W}_{2}}=\frac{{P}_{1}}{{P}_{2}}``
`` ``
`` \Rightarrow \Delta {W}_{1}<\Delta {W}_{2}\left[\because {P}_{2}>{P}_{1}\right]``
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Qstn #9
A gas is contained in a metallic cylinder fitted with a piston. The piston is suddenly moved in to compress the gas and is maintained at this position. As time passes the pressure of the gas in the cylinder
(a) increases
(b) decreases
(c) remains constant
(d) increases or decreases depending on the nature of the gas.
digAnsr: b
Ans : (b) decreases
As the piston of a metallic cylinder containing gas is moved to compress the gas, the volume in which the gas is contained reduces, leading to increase in pressure and temperature. When the time elapses, the heat generated radiates through the metallic cylinder as metals are good conductors of heat. Consequently, the pressure of the gas in the cylinder decreases because of decrease in the temperature.
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#
Section : iii

Qstn #1
The pressure p and volume V of an ideal gas both increase in a process.
(a) Such a process is not possible.
(b) The work done by the system is positive.
(c) The temperature of the system must increase.
(d) Heat supplied to the gas is equal to the change in internal energy.
digAnsr: b,c
Ans :
(b) The work done by the system is positive.

(c) The temperature of the system must increase.
Work done by the system depends on the change in volume during the process `` (\Delta W=p\Delta V)``, where p be the pressure and `` ∆``V be the change in volume of the ideal gas. Since in this process the volume is continuously increasing, the work done by the system will be positive.
According to the ideal gas equation, `` pV=nRT``.
Since both p and V are increasing, temperature (T ) must also increase.
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Qstn #2
In a process on a system, the initial pressure and volume are equal to the final pressure and volume.
(a) The initial temperature must be equal to the final temperature.
(b) The initial internal energy must be equal to the final internal energy.
(c) The net heat given to the system in the process must be zero.
(d) The net work done by the system in the process must be zero.
digAnsr: a,b
Ans : (a) The initial temperature must be equal to the final temperature.
(b) The initial internal energy must be equal to the final internal energy.
a) Let initial pressure, volume and temperature be P₁, V₁ and T₁ and final pressure, volume and temperature be P₂, V₂ and T₂. Then,
`` \frac{{P}_{1}{V}_{1}}{{T}_{1}}=\frac{{P}_{2}{V}_{2}}{{T}_{2}}``
`` ``
`` \Rightarrow {T}_{1}={T}_{2}\left[{P}_{1}={P}_{2}\,\mathrm{\,and\,}{V}_{1}={V}_{2}\right]``
b) Internal energy is given by ΔU = nC_vΔT.
Since ΔT = 0, ΔU = 0.
c) This may not be true because the system may be isothermal and in between, heat may have been given to the system. Also, the system may have done mechanical expansion work and returned back to its original state.
d) Not necessary because an isothermal system may have done work absorbing heat from outside and coming back to the same state losing heat.
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