Thermodynamics

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Questions:

Generic 2

5 x 2 = 10

Can the temperature of a system be increased without heating it?

How is the efficiency of a Carnot engine affected by the nature of the working substance?

What is the change in internal energy of an ideal gas which is compressed/ expanded isothennally? Why?

What amount of heat must be supplied to 2.0 × 10^–2 kg of nitrogen (at room temperature) to raise its temperature by 45 °C at constant pressure ? (Molecular mass of N₂ = 28, R = 8.3 J mol^-1K^-1)

How an adiabatic can be carried practically?

Generic 5

2 x 5 = 10

Two cylinders A and B of equal capacity are connected to each other via a stopcock. A contains a gas at standard temperature and pressure. B is completely evacuated. The entire system in thermally insulated. The stopcock is suddenly opened. Answer the following
(a) What is the change in internal energy of the gas?
(b) What is the change in temperature of the gas?
(c) Do the intermediate states of the system (before settling to the final equilibrium state) lie on its p - V - T surface?

Three moles of an ideal gas kept at constant temperature of 300 K are compressed from a volume of 4L to 1L.Calculate the work done in the process.Take R as 8.31 J/mol-K.

Generic 3

5 x 3 = 15

Tajender was going to Agra with his father in the month of June. It was too hot to tolerate that day. His father was driving the car. He stopped on a petrol pump in the way and got filled the tank of the petrol in the car. He also checked the air in the tyres of the car at air pump. He asked the worker who was there to fill the air that fill the air in the tyre lesser than the normal air. Tajender immediately asked the reason behind it. His father explained that while driving, the air in the tyres expand due to the heat produced by the friction between the road and tyres. Tajender got the answer and became happy.
(i) What values Tajender exhibit?
(ii) Can you design a heat Engine of 100% efficiency? Explain your answer.

Anoop who is the student of class VIII went to a village in Rajasthan with his elder brother a science graduate. It was a month of June. He realised that during the day time, it was extreme hot but during the night it was too cold. He asked his elder brother the reason behind it. Anoop did not feel like it at his residence in Delhi. His brother explained him about Newton's cooling law that in desert places, the sand becomes too hot during day time and according to Newton's Law of cooling "The rate of heating is equal to rate of cooling". Anoop understood this reason very well and became happy.
(i) What qualities, Anoop possess?
(ii) The climate of a harbour town is more temperate than that of a town in a desert at the same altitude. Why

10.

The volume of an ideal gas is V at a pressure P. On increasing the pressure by ΔP, the change in volume of the gas is (ΔV₁) under isothermal conditions and (ΔV₂) under adiabatic conditions. Is ΔV₁ > ΔV₂ or vice-versa and why?

11.

Consider that an ideal gas (n moles) is expanding in a process given by p = f(V), which passes through a point (V_o,P_o ). Show that the gas is absorbing heat at (P₀,V₀), if the slope of the curve p = f(V) is larger than the slope of the adiabat passing through (P_o,V₀).

12.

A Carnot engine absorbs 6 x 10⁵ cal at 227 ^o C. Calculate work done per cycle by the engine if its sink is maintained at 127^o C.

Multiple Choice Question

7 x 1 = 7

13.

An ideal gas heat engine operates in a Carnot cycle between 227°C and 127°C.It absorbs 6 k cal of heat at higher temperature. The amount of heat in k cal rejected to sink is _______.

(a)

4.8

(b)

2.4

(c)

1.2

(d)

6.0

14.

An engine has an efficiency of 1/6 when the temperature of sink is reduced by 62°C, its efficiency is doubled, temperature of the source is _______.

(a)

37°C

(b)

62°C

(c)

99°C

(d)

124°C

15.

The given quantity of an ideal gas is at pressure P and absolute Temperature T. The isothermal bulk Modulus of the gas is _______.

(a)

2/3P

(b)

(c)

3/2P

(d)

16.

For a gas, r = 1.4 then atomicity, C_P, and C_V of the gas are _______.

(a)

Monoatomic 5/2 R, 3/2R

(b)

Monoatomic 7/2 R, 5/2R

(c)

Diatomic7/2 R, 5/2R

(d)

Triatomic 7/2 R, 5/2R

17.

An ideal heat engine exhosting heat at 27°C is to have 25%efficiency. It must take heat at: _______.

(a)

127°C

(b)

227°C

(c)

327°C

(d)

673°C

18.

In an adiabatic change, the pressure P and temperature T of a diatomic gas are related by the relation P ∝ T^C where C equals _______.

(a)

5/3

(b)

2/5

(c)

3/5

(d)

7/2

19.

The internal energy of an ideal gas depends on: _______.

(a)

Pressure

(b)

Volume

(c)

Temperature

(d)

Sizeof molecules

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Answers Key:

Generic 2

5 x 2 = 10

Yes, for example in adiabatic compression

The efficiency is independent of the nature of the working substance.

Zero, because for an ideal gas internal energy is wholly kinetic and it is a function of temperature. As temperature remains constant in an isothermal process, hence, internal energy of an ideal gas remains constant.

Here, mass of gas, m = 2\(\times \)10^-2 kg = 20g
Rise in temperature, \(\Delta \)T = 45^oC
Heat required,\(\Delta \)Q = ?
Molecular mass, M =28
Number of moles, n =\(\frac { m }{ n } =\frac { 20 }{ 28 } =0.714\)
As nitrogen is a diatomic gas, molar specific heat at constant pressure is
\(C_{ p }=\frac { 7 }{ 2 } R=\frac { 7 }{ 2 } \times 8.3J\quad mol^{ -1 }K^{ -1 }\)
\(\\ As\ \Delta Q=nC_{ p }\Delta T\)
\(\\ \therefore \ \Delta Q=0.714\times \frac { 7 }{ 2 } \times 8.3\times 45J=933.4\ J\)

For an adiabatic process, \(\triangle Q=0\) .So, if a process is carried very fast so that heat cannot transferred from system to surroundings and vice-versa, it is an adiabatic process.

Generic 5

2 x 5 = 10

(a). As there is no work done by or on the system, also there is no heat interaction, therefore internal energy of the system remains the same.
Change in internal energy, \(\triangle \)U = 0, as work is done on or by the gas.
(b). The expansion of the gas is not due to any external work done on the system and hence, temperature of the system will not change.
(c). Free expansion is a very fast process and it cannot be controlled. As the intermediate states are non-equilibrium states, hence they can’t be on the P−V−T surface of the system.

\(Given,\mu =3,\ T=300K,\ { V }_{ i }=4\ L,\ { V }_{ f }=1L,\)
\(\\ R=8.31\ J/mol-K,\ W=?\)
\(Work\ done\ in\ isothermal\ process\ is\ given\ by\)
\(\\ W=2.303\mu RT\ log\frac { { V }_{ f } }{ { V }_{ i } } \)
\(\\ =2.303\times 3\times 8.31\times 300\ log\frac { 1 }{ 4 } =-1.037\times { 10 }^{ 4 }J\)

Generic 3

5 x 3 = 15

(i) The values are: possessiveness, keen observer, sharp mind and intelligence.
(ii) The efficiency of a heat engine is
\(η=1-{T_2\over T_1}\)
The efficiency of heat engine will be 100% or 1 if
T₂ = 0K
Since temperature equal to 0K can not be reached, so a heat engine cannot have 100 % efficiency.

(i) Anoop possesses the qualities like having scientific attitude, awareness, intelligence and keen observer.
(ii) The relative humidity in a harbour town is more than that in a town in a desert. Hence the climate of a harbour town is more temperate than that of a town in a desert.

10.

Under isothermal conditions, \(K_i={ΔP\over ΔV_1/V}=P\)
under adiabatic condition, \(K_a={ΔP\over ΔV_2/V}=γP\)
Dividing (ii) by (i), we get
\({ΔV_1\over ΔV_2}=⋎.\ As\ ⋎>1\)
(ΔV₁) > (ΔV₂)

11.

Slope of p = f(V), curve at(V₀,P₀) = f(V₀)
slope of adiabat at (V₀,P₀)
\(=K(-\gamma ){ V }_{ 0 }^{ -1-\gamma }=-\gamma { P }_{ 0 }/{ V }_{ 0 }\)
Now, heat absorbed in the process p = f(V)
dQ = dU + dW
= nC_vdT + pdV
Since, T = (1/nR)pV = (1/nR)Vf(V)
dT = (1/nR)[f(V) + Vf'(V)]dV
Thus, \(\frac { dQ }{ dV } { | }_{ v={ v }_{ 0 } }=\frac { { C }_{ v } }{ R } [f({ V }_{ 0 })+{ V }_{ 0 }f^{ ' }(V_{ 0 })]+f({ V }_{ 0 })\)
\(=[\frac { 1 }{ \gamma -1 } +1]f({ V }_{ 0 })+\frac { { V }_{ 0 }{ f }^{ ' }({ V }_{ 0 }) }{ \gamma -1 }\)
\( \\ =\frac { \gamma }{ \gamma -1 } { P }_{ 0 }+\frac { { V }_{ 0 } }{ \gamma -1 } { f }^{ ' }({ V }_{ 0 })\)
Heat is absorbed when dQ/dV > 0 when gas expands, that is when
\(\gamma { P }_{ 0 }+{ V }_{ 0 }{ f }^{ ' }({ V }_{ 0 })>0\)
\(\\ { f }^{ ' }({ V }_{ 0 })>1-\gamma { P }_{ 0 }/{ v }_{ 0 }\)

12.

Here, heat abs or bed \(=\mathrm{Q}_1=6 \times 10^5 \mathrm{cal}\).
Initial temperature \(=\mathrm{T}_1=227^{\circ} \mathrm{C}=227+273=500 \mathrm{~K}\).
Final temperature \(=\mathrm{T}_2=127^{\circ} \mathrm{C}=127+273=400 \mathrm{~K}\).
As, for Carnot engine;
\( \frac{Q_2}{Q_1}=\frac{T_2}{T_1} \)
\(Q_2=Q_1 \frac{T_2}{T_1} \)
\( \mathrm{Q}_2=\frac{400}{500} \times 6 \times 10^5 \)
\( \mathrm{Q}_2=4.8 \times 10^5 \mathrm{cal} \)
\( \mathrm{Q}_2=\text { Final heat emitted } \)
\( \text { As } \mathrm{w}=\mathrm{Q}_1-\mathrm{Q}_2=6 \times 10^5-4.8 \times 10^5 \)
\(=1.2 \times 10^5 \mathrm{cal} \)
\( \text { Work }=\mathrm{w}=1.2 \times 10^5 \times 4.2 \mathrm{~J} \)
\(\text { Dore }=5.04 \times 10^5 \mathrm{~J} \)