Laws Of Thermodynamics

No, the internal energy of the system necessarily should not increase if heat is added to it.

Explanation

1. Internal energy is defined as the sum of the kinetic energy and potential energy of molecules of the system. It includes only the energy associated with the random motion of molecules of the system.

2. Random motion of molecules is associated with the temperature of the system. Thus, any change in temperature will change the internal energy of the system.

3. Change in internal energy is given as

ΔU=C_v ΔT

Where ΔU= change in internal energy

C_v=molar specific heat at constant volume

ΔT= change in temperature.

4. If ΔT =0 then, ΔU will also be zero.

5. In the isothermal process, where the change in temperature is zero, ΔU is also zero.

6. According to First law of thermodynamics,

ΔQ=ΔU+ΔW

Where ΔQ=heat supplied to the system

ΔU=change in internal energy

ΔW=work done by the system

7. Since for an isothermal process ΔT=0, therefore ΔU=0. So, from the first law of thermodynamics ΔQ=ΔW.

8. For an isothermal process, heat supplied to the system is used up entirely in doing work.

No, the internal energy of a system will not necessarily increase if its temperature is increased.

Explanation

1. Internal energy is defined as the sum of the kinetic energy and potential energy of molecules of the system. It includes only the energy associated with the random motion of molecules of the system.

2. The important thing about internal energy is that it depends only on the state of the system, not how that state was achieved.

3. Thus, the internal energy of a given mass of gas depends on its state described by specifics values of pressure, volume, and temperature.

4. The internal energy of ideal gas depends on only on temperature.

5. For system other than ideal gas, internal energy depends on pressure, volume and temperature combined.

1. Work done on the gas will be zero.

Explanation

Atmospheric pressure decreases, as we increase the height.

So, when the cylinder is lifted from the first floor to second-floor atmospheric pressure on the cylinder will decrease. Hence, the volume of gas inside the cylinder will increase. In the expansion process, work is always done by the gas. So, in such case, no work will be done on the gas. But gas will do work.

2. Work done by the gas will be PΔV.

Explanation

When the cylinder is lifted from the first floor to second-floor atmospheric pressure on the cylinder will decrease. Hence the volume of gas inside the cylinder will increase. This means gas expands. In the expansion, process work is done by the gas.

We know that,

Work done = force ×displacement

Volume = area ×displacement

Therefore,

Work done=pressure ×volume

Let change in the volume of gas = ΔV

The pressure at which gas expands =P

Thus, work done by the gas W

W=PΔV

3. Change in internal energy and temperature will depend on the walls of the container, whether they are insulating or conducting. So, we cannot comment about internal energy and temperature unless we know the nature of the wall.

1. Work done by the force on the block is Fd.

Explanation

Given

Force =F

Displacement due to force = d

We know that

Work done= force × displacement × cos θ

θ in our case is zero as displacement is in the direction of the force.

Work done by the force = F ×d ×cos 0

=F ×d (∵ cos0=1)

∴ Work done by the force on the block is F ×d.

2. Internal energy will not change because of work done by the force on the block.

Explanation

Internal energy is defined as the sum of the kinetic energy and potential energy of molecules of the system. It includes only the energy associated with the random motion of molecules of the system.

If the block moves as a whole system with some velocity, the kinetic energy of the box is not to be included in internal energy.

Energy transferred to the gas is heat energy.

Explanation

Since the outer surface of the gas is rubbed and not displaced/shaken, no work will be done on the gas.

We know that,

Work done = force ×displacement

Force =pressure × area

∴work done = pressure × area × displacement

So, work will only be done when there is either displacement or the gas is expanded/compressed.

The cylinder and gas become warm because of the heat generated by friction between the polishing machine and surface of the cylinder. This heat will warm up the gas and outer surface of the cylinder.

Our hands become warm when we rub them due to the friction between our hands. Friction between our hands supplies heat to the hands.

Work is done when there is displacement in the body.

Work done = force ×displacement

As the bottle is shaken vigorously, that means the bottle is displaced from its position. And this displacement is due to external force. Therefore, work is done on the bottle. Due to this work, the temperature of the liquid increases. No external heat is supplied to the bottle and liquid.

Work done by the system is neither necessarily zero nor nonzero when the final volume is equal to the initial volume.

Explanation

We know that,

Work done = force ×displacement

Volume = area ×displacement

Therefore,

Work done=pressure ×volume

Let change in the volume of system = ΔV = V₂-V₁

Pressure =P

Thus, work done by the system W at constant pressure

W=PΔV

But if V₂=V₁ then, ΔV=0

Work done by the system W=0

So, in an isobaric process (where pressure remains constant) work done by the system will be zero if initial and final volume are equal.

But we also know that in the cyclic process the system returns to its initial state. But still work done is not zero. In fact, in the cyclic process since the system returns to its initial state, internal energy becomes zero. This is because internal energy is a state variable. It depends on initial and final state only. And if initial and final state becomes equal, then change in internal energy will be zero.

According to First law of thermodynamics,

ΔQ=ΔU+ΔW

Where ΔQ=heat supplied to the system

ΔU=change in internal energy

ΔW=work done by the system

In cyclic process ΔU=0.

So, from the first law of thermodynamics ΔQ=ΔW i.e. heat supplied to the system is converted entirely into work in a cyclic process.

Yes, work can be done by a system without changing its volume if the process is cyclic.

Explanation

1. In cyclic process, the system returns to its initial state. Therefore, initial volume becomes equal to final volume and hence ΔV=0.

2. Also, since system returns to initial state, the internal energy of the system remains same.

3. This is because internal energy is a state variable. It depends on initial and final state only. And if initial and final state becomes equal, then change in internal energy will be zero.

4. So, for cyclic process ΔV=0 and ΔU=0.

5. According to First law of thermodynamics,

ΔQ=ΔU+ΔW

Where ΔQ=heat supplied to the system

ΔU=change in internal energy

ΔW=work done by the system

6. So, from the first law of thermodynamics ΔQ=ΔW i.e. heat supplied to the system is converted entirely into work in a cyclic process.

7. Therefore, work can be done by a system without changing its volume.

Yes, the internal energy will also be doubled.

Explanation

1. Internal energy for an ideal gas is given as

U=nC_vT

Where U= internal energy

N=number of moles

C_v=molar specific heat at constant volume

T= temperature

2. Since the ideal gas is continuously pumped into rigid container number of moles are also increasing.

3. When the pressure becomes doubled, the number of moles also gets doubled.

4. It is given that the temperature remains constant. So internal energy depends only on the number of moles.

5. Internal energy will also be doubled as the number of moles is getting doubled.

We know that pressure inside the tyre is greater than the atmospheric pressure of surroundings. So, when a tyre bursts there is an adiabatic expansion of the air (adiabatic because no heat is either supplied or released during expansion). In the expansion process, work is done by the gas. Now work done in an adiabatic process is given as

Where =number of moles

R=gas constant

T₁=initial temperature

T₂=final temperature

= ratio of specific heat at constant pressure and volume

Since work is done by the gas in adiabatic expansion, therefore, W>0. From equation (i), for W>0, T₁>T₂. This means that air

coming out of tyre will be at a lower temperature than when it was inside the tyre.

1. We know that,

Work done = force ×displacement

Volume = area ×displacement

Therefore,

Work done=pressure ×volume

Let change in the volume of object = ΔV = V₂-V₁

Pressure =P

Thus, work done by the object W

W=PΔV

During expansion volume increases. Hence work done will be positive and equals PΔV.

2. According to First law of thermodynamics,

ΔQ=ΔU+ΔW

Where ΔQ=heat supplied to the system

ΔU=change in internal energy

ΔW=work done by the system

Since work done by the gas is not zero, heat supplied to the object will not be equal to internal energy only.

No, it is not a reversible process.

Explanation

1. When we stir a liquid vigorously, it becomes warm because we do work on the liquid and it increases its temperature.

2. To make it a reversible process, would require it to bring the temperature to its initial value by stirring the liquid in opposite direction. Only then we can call it a reversible process.

3. But we know that this is not possible. The only way we can decrease the temperature is by extracting heat from the liquid and not by stirring it in opposite direction.

We know that efficiency of a Carnot engine is

Where W=work done by the engine

Q₁=heat absorbed by the engine

Q₂=heat released by the engine

As seen from the formula (i), efficiency will be 1 when:

i) W=Q₁

ii) Q₂=0

1. The entropy of the system is the measure of molecular disorder or randomness, of a system.

2. When heat is withdrawn from the system, temperature decreases. This means that there is less randomness in the system. So, entropy will decrease.

3. But the heat withdrawn from the system is supplied to surroundings. Therefore, the entropy of the surroundings will increase.

4. The second law of thermodynamics states that net entropy of the universe always increases.

5. So, if the entropy of the system decreases, the entropy of the surroundings increases. Therefore, there is always a net increase in entropy.

6. Hence second law of thermodynamic is not violated in this case.

According to First law of thermodynamics,

ΔQ=ΔU+ΔW

Where ΔQ=heat supplied to the system

ΔU=change in internal energy

ΔW=work done by the system

It is a general law of conservation of energy applicable to any systemin which the energy transfer from or to surroundings is considered.

According to First law of thermodynamics,

ΔQ=ΔU+ΔW

Where ΔQ=heat supplied to the system

ΔU=change in internal energy

ΔW=work done by the system

For an isothermal process ΔT=0, therefore ΔU=0.

Then,

ΔQ=ΔW

When heat is supplied to the system gas expands i.e. the volume of the gas increases.

We know that,

Work done = force ×displacement

Volume = area ×displacement

Therefore,

Work done=pressure ×volume

Let change in the volume of object = ΔV = V₂-V₁

Pressure =P

Thus, work done by the object W

W=PΔV

During expansion volume increases. Hence, work done will be positive and equals PV.

Initial and final points of both processes A and B are same. Therefore, internal energy in both the processes will be the same because internal energy is a state variable, independent of the path taken.

The area under the P-V curve gives the work done on the system. From the graph, it can be seen the area under the curve for process A is more than the area under the curve for process B., therefore, work done on the system in process A ΔW₁ is more than work done on the system in process B ΔW₂

ΔW₁ > ΔW₂ …..(i)

According to First law of thermodynamics,

ΔQ=ΔU+ΔW

Where ΔQ=heat supplied/extracted to/from the system

ΔU=change in internal energy

ΔW=work done by/on the system

For process A

ΔQ₁=ΔU+ΔW₁ …(ii)

For process B

ΔQ₂=ΔU+ΔW₂ …(iii)

From equation (i) ,(ii) and (iii) it is clear that ΔQ₁ > ΔQ_2.

Initial and final points of both processes A and B are same. Therefore, change in internal energy in both the processes will be the same because internal energy is a state variable, independent of the path taken.

Therefore, ΔU₁=ΔU₂=0.

We know that,

Work done = force ×displacement

Volume = area ×displacement

Therefore,

Work done=pressure volume

Let change in the volume of system = ΔV = V₂-V₁

Pressure =P

Thus, work done by the system W at constant pressure

W=PΔV

From the graph we can see that V₂ > V₁ i.e. final volume is greater than the initial volume.

So, work done by the system continuously increases.

Statement A: For an isothermal process (where the temperature remains constant) if heat is added to a system temperature will not increases. So, statement A is wrong.

Statement B: We know that,

Work done = force ×displacement

Volume = area ×displacement

Therefore,

Work done=pressure ×volume

Let change in the volume of system = ΔV = V₂-V₁

Pressure =P

Thus, work done by the system W at constant pressure

W=PΔV

W>0 and positive only when V₂ > V₁.

Therefore, statement B is correct.

Since the graph between P and T is a straight line passing through the origin, therefore PT.

P can only be proportional to T when the volume is kept constant. This can be easily proved from the ideal gas equation which is

PV=RT

Since R is already a constant, if volume also becomes constant then PT.

Constant volume implies ΔV=0.

We know that,

Work done = force ×displacement

Volume = area ×displacement

Therefore,

Work done=pressure ×volume

Let change in the volume of system = ΔV = V₂-V₁

Pressure =P

Thus, work done by the system W

W=PΔV

For ΔV=0, W=0.

Given V₁=V₂

We know that,

Work done = force ×displacement

Volume = area ×displacement

Therefore,

Work done=pressure ×volume

Let change in the volume of system = ΔV = V₂-V₁

Pressure =P

Thus, work done by the system W

ΔW=PΔV

For process A ΔW₁=P₁ΔV₁

For process B ΔW₂=P₂ΔV₂

Since ΔV₁=ΔV₂ we can write,

P₁ < P₂ (from graph)

Therefore ΔW₁ < ΔW₂.

1. When the piston is moved to compress the gas, the volume of the gas reduces and pressure and temperature increases.

2. After that piston is maintained at this position. So, volume becomes constant.

3. But during this sudden compression heat is generated. And since the cylinder is metallic this heat can escape out of the cylinder as metals are good conductors of heat.

4. When heat escapes out of the cylinder, the temperature of the gas will decrease which will, in turn, decrease the pressure of the gas.

We know that,

Work done = force ×displacement

Volume = area ×displacement

Therefore,

Work done=pressure ×volume

Let change in the volume of system = ΔV = V₂-V₁

Pressure =P

Thus, work done by the system W

ΔW=PΔV

It is given that volume in increasing i.e. ΔV>0. So, ΔW>0 and positive.

From ideal gas equation, we know that

PV=nRT

Where P=pressure

V=volume

n=number of moles

R=gas constant

T=temperature

So, if both pressure and volume are increasing, then

the temperature must also increase as n and R are constant.

From ideal gas equation, we know that

PV=nRT

Where P=pressure

V=volume

n=number of moles

R=gas constant

T=temperature

If P_iV_i=P_fV_f

Then nRT_i=nRT_f

∴ T_i=T_f

Since the initial state of the system (P_i, V_i, T_i) is equal to final state of the system (P_f, V_f, T_f), initial and final internal energy will also be equal as internal energy is a state variable.

According to First law of thermodynamics,

ΔQ=ΔU+ΔW

Where ΔQ=heat supplied/extracted to/from the system

ΔU=change in internal energy

ΔW=work done by/on the system

∴ ΔU=ΔQ – ΔW

It is given that initial state (p₁, V₁) and final state (p₂, V₂) is same for both the method.

So, change in internal energy will be the same for both the methods as internal energy is a state variable and is independent of the path taken to achieve a state.

As seen from the figure initial state of A is the final state of B and vice-versa.

So, the change in internal energy of process A and B will be equal in magnitude but with a negative sign.

So ΔU₁=-ΔU₂

∴ ΔU₁+ΔU₂=0.

1. The internal energy of an ideal gas is proportional to the temperature. Since internal energy is decreasing, therefore, the temperature must decrease.

2. According to the question, change in internal energy is equal to change in work done.

3. According to First law of thermodynamics,

ΔQ=ΔU+ΔW

Where ΔQ=heat supplied/extracted to/from the system

ΔU=change in internal energy

ΔW=work done by/on the system

4. We know for adiabatic process ΔQ=0. So, from first law ΔU=ΔW.

Given

Initial temperature of water T₁=15℃ =288K

Final temperature of water T₂=17℃ =290K

Specific heat capacity of copper c_c=420 J kg^–1 K^–1

Specific heat capacity of water c_w=4200 J kg^–1 K^–1

Mass of copper vessel m_c=100g = 100×10^-3kg

Mass of water m_w=200g = 200×10^-3kg

a) It is given that copper vessel is thermally insulated. Therefore, no heat from the surroundings can be transferred to the liquid-vessel system.

b) Work done on this system will be

ΔW=m_wc_wΔT + m_cc_cΔT

ΔT=T₂-T₁=290-288=2K

So,

ΔW= 200×10^-3×4200×2 + 100×10^-3×420×2

ΔW=1680 + 84=1764J

∴ work done on the system is 1764J.

c) From first law of thermodynamics, we know that,

ΔQ=ΔU+ΔW

Where ΔQ=heat supplied to the system

ΔU=change in internal energy

ΔW=work done by the system

From part (a) we have concluded that ΔQ=0

So, first law becomes

ΔU=-ΔW

Work done on the system ΔW=1764 J

So, work done by the system ΔW=-1764J

ΔU = -(-1764) = 1764J

∴ Increase in internal energy of the system = 1764J.

Given

Mass attaches to the pulley m = 12kg

Heat capacity of liquid s =4200 JK^-1

Height through which mass fall= 70cm=0.7m

a) Paddle immersed in liquid is kept in an adiabatic container. So, no heat can be either supplied or extracted to the liquid. Therefore, heat given to liquid is zero.

b) As no heat is supplied to liquid and pulley is frictionless, work done on the liquid will be equal to the potential energy of the mass attached to the pulley.

Work done on the liquid = potential energy of mass

Potential energy of mass= mgh

Where g=acceleration due to gravity=10ms^-2

Potential energy of mass= 12×10×0.7=84J

∴ work done on liquid =84J.

c) The mechanical work calculated in the second part will be converted into heat. This heat will be supplied to liquid and due to which temperature of the liquid will rise.

We know that,

Heat capacity

Where ΔQ = heat supplied

ΔT=rise in temperature

Since work done is equal to heat supplied, therefore

∴ the rise in temperature of the liquid will be 0.02K.

Given

Mass of block m=100kg

Initial velocity u=2 m/s

Coefficient of kinetic friction μ = 0.20

a) Since the block comes to stop final velocity will be zero.

So, final velocity v=0 m/s

When the block is moving over belt there is kinetic friction between the lower surface of the block and upper surface of the belt. And we know that heat is produced due to friction between two surfaces. Now because of this heat, the internal energy of block will change.

So,

Kinetic energy lost in heat due to friction = change in the internal energy

Kinetic energy lost = initial kinetic energy- final kinetic energy

change in internal energy is 200J.

b) Given

The velocity of the frame of reference u_o= 2m/s

So, in this frame of reference initial and final velocity of the block will change.

New initial velocity u’=u-u_o =2-2 = 0m/s

New final velocity v’ = v-u_o =0-2 =-2m/s

So, increase in kinetic energy = (final – initial) kinetic energy

Increase in kinetic energy in com frame of reference is 200J.

c) Total work done in com frame of reference will be work done due to friction plus work done to give final velocity.

We know that force of friction f=N

Where = coefficient of friction =0.02

N=normal reaction =mg

f = 0.02×100×10 = 200N

from newton’s second law of motion

force= mass ×acceleration

so,

200=100×acceleration

Using the third equation of motion,

v’²-u’²=2as

where s=displacement as seen in com frame of reference

(-2)²-0=2×2×s

s=1m

work done due to friction W_f =force ×displacement

W_f=200×1=200J

Now to calculate work done to give final velocity, we will work-energy theorem.

According to work-energy theorem,

Work done = change in kinetic energy

So,

Total work done W=W’ + W_f =200+200=400J_.

∴ work done in com frame of reference= 400J.

Given

Heat supplied ΔQ = 100J

From first law of thermodynamics, we know that,

ΔQ=ΔU+ΔW

Where ΔQ=heat supplied to the system

ΔU=change in internal energy

ΔW=work done by the system

We know that,

Work done = force ×displacement

Volume = areadisplacement

Therefore,

Work done=pressure ×volume

Let change in the volume of system = ΔV = V₂-V₁

Pressure =P

Thus, work done by the gas

ΔW=PΔV

Since the gas is kept in a rigid container, therefore ΔV=0 in this case.

So, ΔW=0

Thus, first law will become

ΔQ=ΔU = 100J

∴ change in internal energy will be 100J.

Given

Initial pressure P₁=10kPa=10×10³Pa

Final pressure P₂=50kPa=50×10³Pa

Initial volume V₁=200cc=200×10^-6m³

Final volume V₂=50cc= 50×10^-6m³

a) We know that,

Work done = force ×displacement

Volume = area ×displacement

Therefore,

Work done=pressure ×volume

Let change in the volume of system = ΔV = V₂-V₁

Pressure =P

Thus, work done by the gas

ΔW=PΔV

Here we have given two values of pressure. So, we will take the average value of pressure

Average pressure P

Therefore, ΔW = 30×10³× (50-200)×10^-6

ΔW= -4.5J

Work done by the gas is -4.5J.

b) Given that no heat is supplied or extracted from the gas.

From first law of thermodynamics, we know that,

ΔQ=ΔU+ΔW

Where ΔQ=heat supplied to the system

ΔU=change in internal energy

ΔW=work done by the system

Since ΔQ=0

Therefore ΔU = -ΔW = -(-4.5) J=4.5J

∴ the change in internal energy of the gas is 4.5J.

Given

Initial pressure P_i

Final pressure P_f

Initial temperature T_i

Final temperature T_f

It is given that

From ideal gas equation, we know

PV=nRT

Where V=volume of gas

R=gas constant

n= number of moles

applying ideal gas equation for both processes, we get

And

From equation (i), (ii) and (iii), we get

∴ V_i=V_f

We know that,

Work done = force ×displacement

Volume = area ×displacement

Therefore,

Work done=pressure × volume

Let change in the volume of system = ΔV = V_f-V_i

Pressure =P

Thus, work done by the gas

ΔW=PΔV=P (V_f-V_i) =0

∴ work done by the gas is zero.

We know that work done by the gas is given as

ΔW=PΔV

From graph we can write

V_A =V_C = 10cc=10×10^-6m³

V_D=V_B = 25cc= 25×10^-6m³

P_B=P_C=30kPa= 30×10³Pa

P_A=P_D=10kPa=10×10³Pa

Work done in path ADB W_ADB=W_AD + W_DB

W_ADB = P_A (V_D-V_A) + 0 (∵ W_DB =0 because V_D=V_B)

=10×10³×(25-10)×10^-6

=0.15J

Work done in path AB W_AB= P_avg(V_B-V_A)

W_AB=20×10³× (25-10)×10^-6

=0.30J

Work done in path ACB W_ACB= W_AC+W_BC

W_ACB = 0+P_B(V_B-V_C) (∵ W_AC=0 because V_A=V_C)

=30×10³×(25-10)×10^-6

=0.45J

Given

Heat absorbed in process abc ΔQ₁=80J

Work done by the system in process abc ΔW₁=30J

Work done by the system in process adc ΔW₂=10J

Let heat absorbed into the system during process adc =ΔQ₂

Now initial point a and final point c is the same for both the processes is the same. So, change in internal energy will be the same for both the process, as internal energy is a state function independent of the path taken.

Therefore,

ΔU₁=ΔU₂=ΔU ……(i)

From first law of thermodynamics, we know that,

ΔQ=ΔU+ΔW

Where ΔQ=heat supplied to the system

ΔU=change in internal energy

ΔW=work done by the system

Using first law of thermodynamics for process abc

ΔQ₁=ΔU₁+ΔW₁

ΔU₁=ΔQ₁-ΔW₁=80-30=50J

Using first law of thermodynamics for process adc

ΔQ₂=ΔU₂+ΔW₂

=ΔU₁+ΔW₂ (from (i))

= 50+10=60J

∴ heat absorbed into the system during process adc= 60J.

From graph we can write

V_A =V_D = 200cc=200×10^-6m³

V_B=V_C = 400cc= 400×10^-6m³

P_B=P_D=155kPa= 155×10³Pa

P_A=P_C=50kPa=50×10³Pa

Given

Heat absorbed in process ABC ΔQ₁=50cal= 50×4.2 J=210J

Let heat absorbed into the system during process ADC =ΔQ₂

We know that work done by the gas is given as

ΔW=PΔV

Work done in path ACB W_ACB =ΔW₁= W_AC+W_BC

ΔW₁=P_A(V_C-V_A) + 0 (W_BC=0 because V_B=V_C)

=50×10³×(400-200)×10^-6

=10J

Work done in path ADB W_ADB=ΔW2=W_AD + W_DB

ΔW₂ = P_B (V_D-V_B) +0 (∵ W_AD=0 because V_A=V_D)

=155×10³×(400-200)×10^-6

=31 J

Now initial point A and final point C is the same for both the processes is the same. So, change in internal energy will be the same for both the process, as internal energy is a state function independent of the path taken.

Therefore,

ΔU₁=ΔU₂=ΔU ……(i)

From first law of thermodynamics, we know that,

ΔQ=ΔU+ΔW

Where ΔQ=heat supplied to the system

ΔU=change in internal energy

ΔW=work done by the system

Using first law of thermodynamics for process ABC

ΔQ₁=ΔU₁+ΔW₁

ΔU₁=ΔQ₁-ΔW₁=210-10=200J

Using first law of thermodynamics for process adc

ΔQ₂=ΔU₂+ΔW₂

=ΔU₁+ΔW₂ (from (i))

=200+31=231J.

∴ heat supplied to the system during process ADC=231J.

We know that in the cyclic process the system returns to its initial state. So, change internal energy in the cyclic process will be zero as internal energy is a state function.

From first law of thermodynamics, we know that,

ΔQ=ΔU+ΔW

Where ΔQ=heat supplied to the system

ΔU=change in internal energy

ΔW=work done by the system

Since ΔU=0, first law becomes

ΔQ=ΔW

In a PV graph work done is equal to the area under the curve.

ΔW= area of a circle

ΔQ=ΔW=area of circle

Diameter of the circle =300-100=200

Area of the circle =π × (radius)²

=π × 100×100×10^-6×10³

(10^-6×10³ is because volume and pressure are given in cc and kPa respectively)

Area of circle =3.14×10=31.4

∴ Heat absorbed by a system = 31.4J.

‘J’ is mechanical equivalent of heat a conversion factor between two different units of energy: calorie to joule

From the graph we can write

V_A=V_B=500cc=500×10^-6m³

V_C=700cc=700×10^-6m³

P_A=P_C=100kPa=100×10³ Pa

P_B=200kPa=200×10³Pa

We know that work done by the gas is given as

ΔW=PΔV

Work done in path AB=0 as V_A=V_B.

Work done in path CA ΔW₁=P_A(V_A-V_C)

=100×10³× (500-700)×10^-6

=-20J

Work done in path BC ΔW₂= P_avg(V_C-V_B)

ΔW₂=150×10³×(700-500)×10^-6

=30J

Total work done in process ABCA=ΔW=ΔW₁+ΔW₂

=30-20=10J

We know that in the cyclic process the system returns to its initial state. So, change internal energy in the cyclic process will be zero as internal energy is a state function.

From first law of thermodynamics, we know that,

ΔQ=ΔU+ΔW

Where ΔQ=heat supplied to the system

ΔU=change in internal energy

ΔW=work done by the system

Since ΔU=0, first law becomes

ΔQ=ΔW=10J

But it is given in question that ΔQ=2.4cal

So, 2.4×J=10Joule

∴ value of ‘J’ is 4.17J/cal.

‘J' is mechanical equivalent of heat a conversion factor between two different units of energy: calorie to the joule.

Given

Heat given to system =2625cal =2625×J J

Change in internal energy= 5000J

From graph

V_a=0.02m³

V_b=V_c=0.05m³

P_a=P_b=200kPa=200×10³Pa

P_c=300kPa=300×10³Pa

We know that work done by the gas is given as

ΔW=PΔV

Work done in process abc=ΔW=W_ab+W_bc

ΔW=P_a(V_b-V_a)+0 (∵ W_bc=0 because V_b=V_c)

ΔW=200×10³×(0.05-0.02)=6000J

From first law of thermodynamics, we know that,

ΔQ=ΔU+ΔW

Where ΔQ=heat supplied to the system

ΔU=change in internal energy

ΔW=work done by the system

ΔQ=5000+6000=11000J

But ΔQ=2625cal =2625×J J

Therefore,

2625×J=11000

∴ value of ‘J’ is 4.19joule/cal.

Given

Heat extracted from the system ΔQ=-70cal=-70×4.2=-294J

The negative sign is because heat is extracted from the system.

From graph

V_A=250cc=250×10^-6m³

V_B=100cc=100×10^-6m³

P_A=200kPa=200×10³Pa

P_B=500kPa=500×10³Pa

We know that work done by the gas is given as

ΔW=PΔV

Here since we have two values of pressure we will take average pressure.

ΔW=P_avg(V_A-V_B)

= 350×10³× (250-100) ×10^-6

=-52.5J

From first law of thermodynamics, we know that,

ΔQ=ΔU+ΔW

Where ΔQ=heat supplied to the system

ΔU=change in internal energy

ΔW=work done by the system

Therefore,

ΔU=-294-(-52.5) =-241.5J

∴ change in internal energy is -241.5J.

Given

Constant pressure p=1.0×10⁵ Pa

Change in volume ΔV=(200-100)×10^-6m³=10^-4m³

Internal energy U= 1.5pV

So, change in internal energy ΔU=1.5pΔV

=1.5×1.0×10⁵×10^-4

=15J

We know that work done by the gas is given as

ΔW=pΔV=1.0×10⁵×10^-4=10J

From first law of thermodynamics, we know that,

ΔQ=ΔU+ΔW

Where ΔQ=heat supplied to the system

ΔU=change in internal energy

ΔW=work done by the system

Therefore,

ΔQ=15+10=25J

∴ heat absorbed by the system=25J.

Given

Heat supplied to system ΔQ = 10J

Atmospheric pressure P= 100kPa =100×10³Pa

Displacement of the piston d=10cm=0.1m

Area of cross section of cylinder A=4cm² =410^-4m²

Gas will expand when is heat to the system. Therefore, the volume of gas expanded ΔV

ΔV=area ×displacement

=A ×d

ΔV=4×0.1×10^-4=40×10^-6m³

We know that work done by the gas is given as

ΔW=PΔV

=100×10³×40×10^-6

ΔW =4J

From first law of thermodynamics, we know that,

ΔQ=ΔU+ΔW

Where ΔQ=heat supplied to the system

ΔU=change in internal energy

ΔW=work done by the system

ΔU=ΔQ-ΔW = 10-4 =6J

Thus, the increase in internal energy is 6J.

From graph

V_a=2m³

V_b=V_c=2.5m³

P_a=P_b=100kPa=10010³Pa

P_c=200kPa=20010³Pa

Work done in process ABCA = area enclosed by the triangle ABC

ΔW =0.5×BC×AB

Where BC= height of the triangle

AB= base of triangle

So,

ΔW=0.5×(200-100)×10³(2.5-2)

ΔW=0.5×100×0.5×10³ =25000J

a) Process ABCA is a cyclic process. The system is brought back to its initial state. Since internal energy is a state function, change in internal energy will be zero.

So, ΔU=0

From first law of thermodynamics, we know that,

ΔQ=ΔU+ΔW

Where ΔQ=heat supplied to the system

ΔU=change in internal energy

ΔW=work done by the system

∴ ΔQ=ΔW=25000J

Since work done is positive, work is done by the gas. When work is done by the gas heat is supplied to the system.

b) Amount of heat supplied = work done by the gas in the cyclic process

∴ ΔQ=ΔW=25000J.

Given

Heat extracted from the system ΔQ=-1200J

(negative sign is because heat is extracted from the system)

Number of moles of the gas n=2.0

From the graph we can write

T_A=300K

T_B=500K

V_A=V_C

We know that work done by the gas is given as

ΔW=PΔV

Where ΔV =change in volume

P =pressure

So, work done along line CA will be zero, as V_A=V_C.

Thus, total work done will be

ΔW=W_AB+W_BC

W_AB = P(V_B-V_A)

But we know that ideal gas equation is

PV=nRT

Where n= number of moles

R=gas constant

T=temperature

∴ PΔV=nRΔT

Thus, we can write W_AB = P(V_B-V_A) = nR(T_B-T_A)

∴ ΔW= nR(T_B-T_A) + W_BC

From first law of thermodynamics, we know that,

ΔQ=ΔU+ΔW

Where ΔQ=heat supplied to the system

ΔU=change in internal energy

ΔW=work done by the system

Process ABCA is a cyclic process. The system is brought back to its initial state. Since internal energy is a state function, change in internal energy will be zero.

So, ΔU=0.

So, first law becomes

ΔQ=ΔW

ΔQ= nR(T_B-T_A) + W_BC

We know that R=8.31J/K mol

W_BC =ΔQ- nR(T_B-T_A)

= -1200 - 2×8.31× (500-300)

=-1200 – 3324

=-4524J

Thus, work done along path BC =-4524J

Given

Number of moles n=2

Temperature at b T_b=500K

Temperature at c T_c=300K

From the graph it is clear that

T_b=T_a and T_d=T_c

Thus, path ab and cd are isothermal paths.

We know that work done by the gas is given as

ΔW=PΔV

Where ΔV =change in volume

P =pressure

Again, from the graph we can see that ΔV=0 for path bc and da.

Therefore, work done along path bc and da are zero.

So, total work done ΔW=W_ab+W_cd

We know that work done in an isothermal process is given as

Where n=number of moles

R=gas constant =8.31J/Kmol

T=temperature

V_f=final volume

V_i=initial volume

W_ab = 8310×ln2

Similarly,

W_cd = -4986 ×ln2

So, total work done ΔW=W_ab+W_cd

=8310×ln2 - 4986×ln2

=3324×ln2

ΔW =3324×0.693=2304.02J

From first law of thermodynamics, we know that,

ΔQ=ΔU+ΔW

Where ΔQ=heat supplied to the system

ΔU=change in internal energy

ΔW=work done by the system

Process ABCA is a cyclic process. The system is brought back to its initial state. Since internal energy is a state function, change in internal energy will be zero.

So, ΔU=0.

So, first law becomes

ΔQ=ΔW=2304.02J

Thus, heat absorbed by the system is 2304.02J.

Given

Mass of water m = 2kg

Change in temperature T =4℃ -0℃ =4℃

Specific heat capacity of water c =4200 J kg^–1 K^–1

Density of water at 0°C=999.9 kg m^–3

Density of water at 4°C=1000 kg m^–3

Atmospheric pressure P = 10⁵ Pa.

We know that specific heat capacity is given by

Where ΔQ = heat supplied to the system

Therefore, ΔQ=cmΔT

=4200×2×4=33600J

We know that work done by the gas is given as

ΔW=PΔV

Where ΔV =change in volume

P =pressure

Also,

Volume at 0^oC V₁

Similarly, volume at 4^oC V₂

∴ ΔW=P(V₂-V₁)

From first law of thermodynamics, we know that,

Q=U+ΔW

Where ΔQ=heat supplied to the system

ΔU=change in internal energy

ΔW=work done by the system

ΔU=ΔQ-ΔW

=33600-(-0.02)

=33599.98J

Thus, change in internal energy is 33599.98J.

Given

The density of steam ρ’= 0.6 kg m^–3

Mass of water m=10g =0.010kg

Specific heat capacity of water c = 4200 J kg^–1 °C^–1

latent heat of vaporization of water L = 2.25 × 10⁶ J kg^–1.

Pressure P =100kPa =100×10⁵Pa

Change in temperature ΔT= (100-0) ^oC =100^oC

Density of water ρ =1000 kg m^-3

We know that specific heat capacity is given by

Where ΔQ = heat supplied to the system

Therefore, ΔQ= cmΔT

Also, ΔQ =mL

Where m= mass of the substance

L=latent heat

Therefore, ΔQ=mL + cmΔT

=0.010×2.25 × 10⁶ + 4200×0.01×100

=22500+4200

=26700J

We know that work done by the gas is given as

ΔW=PΔV

Where ΔV =change in volume

P =pressure

Also,

ΔW=10⁵×0.01699=1699 J

From first law of thermodynamics, we know that,

ΔQ=ΔU+ΔW

Where ΔQ=heat supplied to the system

ΔU=change in internal energy

ΔW=work done by the system

ΔU=ΔQ-ΔW

=26700-1699

=25001J

Thus, change in internal energy is 25001J.

a) According to the question, the diathermic separator between both the part is fixed. So, no change in volume will be observed. And as we know work done on gas is PV. Therefore, no work will be done on the left part during the process as the volume is not changing.

First, we will calculate the final temperature and then final pressure.

(c) Given

Pressure of left chamber =P₁

Pressure of right chamber =P₂

Temperature of left chamber =T₁

Temperature of right chamber =T₂

Let the number of moles in the left chamber be n₁

Number of moles in the right chamber be n₂

Diathermic wall has divided the tube in two equal part. So, the volume of the left and the right chamber will be V/2.

Applying ideal gas equation in the left chamber,

Similarly applying ideal gas equation in the right chamber,

Total number of moles n=n₁ +n₂

The internal energy of ideal gas is given as

U=nC_vT

Where C_v=molar specific heat at constant volume

T=temperature.

According to question,

U=1.5nRT

nC_vT=1.5nRT

So, C_v=1.5R

The internal energy of the left chamber U₁=n₁C_vT₁

The internal energy of right chamber U₂= n₂C_vT₂

Total internal energy U= U₁+U₂

1.5nRT= n₁C_vT₁+ n₂C_vT₂

1.5nRT=C_v(n₁T₁+n₂T₂)

Substituting the value of C_v

1.5nRT=1.5R(n₁T₁+n₂T₂)

nT=n₁T₁+n₂T₂

substituting the value of n₁ and n₂ in above equation

Substituting the value of n in the above equation,

Thus, final equilibrium temperature T=.

(b) Now we will find final pressures on both sides.

Let final pressure of left chamber P₁’

Final pressure of right chamber P₂’

Applying ideal gas equation in the left chamber before and after equilibrium

From equation (i) and (ii),

Substituting the value of T,

Applying ideal gas equation in the right chamber before and after equilibrium

From equation (iii) and (iv),

Substituting the value of T,

(d) The internal energy of ideal gas is given as

U=nC_vT

Where C_v=molar specific heat at constant volume

T=temperature.

As stated in part (a) no work will be done on either chamber of the vessel as the diathermic separator is fixed.

So, ΔW=0 for the right chamber of the tube.

From first law of thermodynamics, we know that,

ΔQ=ΔU+ΔW

Where ΔQ=heat supplied to the system

ΔU=change in internal energy

ΔW=work done by the system

ΔQ=ΔU

Change in internal energy of the right chamber after equilibrium has reached will be

ΔU=n₂C_vT₂ -n₂C_vT

Substituting the value of n₂, C_v and T in the above equation

Thus, heat flown from left to right chamber is
.

a) According to the question, conducting separator between both the part is fixed. So, no change in volume will be observed. And as we know work done a gas is PV. Therefore, no work will be done on the left part during the process as the volume is not changing.

b) Given

The vessel is divided into two equal parts.

So, the volume of left part V₁ and volume of right part V₂ will half of the total volume.

Initial pressure on each side is p

Number of moles on left side n₁ =1

Number of moles on right side n₂=2

Let initial temperature for the left and right part be T₁ and T₂ respectively.

For the left part, applying the ideal gas equation

Similarly, for the right part

c) It is given that the internal energy of the gas is

U=1.5nRT

Where n=number of moles

R=gas constant

T=final equilibrium temperature of both parts

The internal energy of ideal gas is given as

U=nC_vT

Where C_v=molar specific heat at constant volume

T=temperature.

Total moles n=n₁+n₂=1+2=3

U=3C_vT

Let U₁ and U₂ be the internal energy of the left and right part respectively.

So, U₁=n₁C_vT₁= C_vT₁

And U₂=n₂C_vT₂=2C_vT₂

(gas is same on both the part so Cv will be same)

Now

U=U₁+U₂

3C_vT = C_vT₁+ 2C_vT₂

3T=T₁+2T₂

d) As stated in part (a) no work will be done on either part of the vessel as conducting separator is fixed.

So, ΔW=0 for the right part of the vessel.

From first law of thermodynamics, we know that,

ΔQ=ΔU+ΔW

Where ΔQ=heat supplied to the system

ΔU=change in internal energy

ΔW=work done by the system

∴ ΔQ=ΔU

It is given that the internal energy of the gas is

U=1.5nRT

Where n=number of moles

R=gas constant

T=final equilibrium temperature of both parts

ΔU=1.5nRΔT

For the right part, change in internal energy after equilibrium has reached will be due to the change in temperature from T₂ to T.

ΔU=1.5n₂R(T-T₂)

Thus, heat given to the right part is .

e) Since ΔW is zero as the volume is fixed, therefore, the first law of thermodynamics.

But since heat is given to the right part that means heat is extracted from the left part. So, the internal energy of the left part of will decrease.

Therefore, heat given to the right part will be equal to the negative internal energy of the left part.

ΔQ=-ΔU (left part)

.