Solved Examples on Thermodynamics

Question:1When a system is taken from state

ito statefalong the pathiafin below figure, it is found thatQ= 50 J andW= -20 J. Along the pathibf,Q= 36 J.(a)What isWalong the pathibf?(b)IfW= +13 J for the curved return pathfi, what isQfor this path?(c)TakeE_{int,i}= 10 J. What isE_{int.f}?(d)IfE_{int,b}= 22 J, findQfor processiband processbf.

Concept: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}So,

W= ΔE_{int }-QAnd

Q= ΔE_{int }–WHere

Qis the energy transferred (as heat) between the system and environment,Wis 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

Qto be positive when heat is transferred into the system andWto be positive when work is done on the system.

Solution:A system is taken from

ito statefalong the pathiaf, which as shown in below figure.

(a) Δ

E_{int}along any path between these two pointsiandfis,Δ

E_{int}=Q+W= (50 J)+(-20 J)

=30 J

To obtain the

Walong the pathibf, substitute 30 J for ΔE_{int}, 36 J forQin the equationW= ΔE_{int }–Q, we get,

W= ΔE_{int }–Q= (30 J)-(36 J) = -6 J

Thus, the work done

Walong the pathibfwould be -6 J.

(b)To obtain the

Qfor curved return pathfi, substitute -30 J for ΔE_{int}and +13 J forWin the equationQ= ΔE_{int }–W, we get,

Q= ΔE_{int }–W=(-30 J)-(+13 J) = - 43 J

Thus the

Qfor curved return pathfiwill be -43 J.

(c)As, Δ

E_{int}=E_{int,f}-E_{int,i}So,

E_{int,f}= ΔE_{int}+E_{int,i}To obtain

E_{int,f}, substitute 10 J forE_{int,i}and 30 J for ΔE_{int}in the equationE_{int,f}= ΔE_{int}+E_{int,i}we get,

E_{int,f}= ΔE_{int}+E_{int,i}= (30 J)+(10 J) = 40 J

Therefore, the value of

E_{int,f}would be 40 J.

(d)Δ

E_{intib}= (22 J) – (10 J) =12 JBut,

Δ

E_{intbf}= (40 J) - (22 J) =18 JAnd there is no work done on the path

bf, since volume is constant.So,

Q= Δ_{bf}E_{intbf}-W_{bf}And

Q=_{ib }Q-_{ibf}Q_{bf}To obtain the value of

Q, substitute 18 J for Δ_{bf}E_{intbf}and 0 J forWin the equationQ= Δ_{bf}E_{intbf}-W, we get,_{bf}

Q= Δ_{bf}E_{intbf}-W= (18 J) – (0 J)= 18 J_{bf }Thus the value of

Qwill be 18 J._{bf}To obtain the value of

Q, substitute 36 J for_{ib}Qand 18 J for_{ibf}Qin the equation_{bf}Q=_{ib }Q-_{ibf}Q, we get,_{bf}

Q=_{ib }Q-_{ibf}Q_{bf}= (36 J) – (18 J) = 18 J

Therefore the value of

Qwould be 18 J._{ib}_____________________________________________________________________________________

Question:2One mole of an ideal monoatomic gas is caused to go through the cycle shown in below figure. (a) How much work is done on the gas in expanding the gas from

atocalong the pathabc?(b)What is the change in internal energy and entropy in going through one complete cycle? Express all answers in terms of the pressurep_{0}and volumeV_{0}at point a in the diagram.

Concept:Work done

Wat constant pressure is defined as,

W=pΔVHere p is the pressure and Δ

Vis the change in volume.Change in internal energy Δ

Eis defined as,Δ

E= 3/2nRΔTHere

nis the number of moles,Ris the gas constant and ΔTis the change in temperature.Change in entropy Δ

Sis defined as,Δ

S= 3/2 ln (T_{f}/T_{i})Here,

T_{f}is the final temperature andT_{i}is the initial temperature.One mole of an ideal monoatomic gas is caused to go through the cycle as shown in the above figure.

Solution:

(a)To obtain the work done

Walong the path_{abc}abc, first we have to find out the work doneWalong the path_{ab}ab.From the figure we observed that,

W=_{ab}pΔV=

p_{0}(V_{0}-4V_{0}) = -3p_{0}V_{0}So,

W=_{abc}W= -3_{ab}p_{0}V_{0}From the above observation we conclude that, the work done on the gas along the path

abcwould be -3p_{0}V_{0}.

(b)The change in internal energy Δ

Ealong the path_{bc}btocwould be,Δ

E= 3/2_{bc}nRΔT_{bc}= 3/2 (

nRT–_{c}nRT)_{b}= 3/2(

p_{c}V_{c }–p_{b}V_{b}) (pV=nRT)= 3/2[(2

p_{0}) (4V_{0}) -(p_{0}) (4V_{0})]= 3/2 [8

p_{0}V_{0}- 4p_{0}V_{0}]= 6p_{0}V_{0}The change in entropy Δ

Salong the path_{bc}btocwould be,Δ

S= 3/2_{bc}nRln(T/_{c}T)_{b}= 3/2

nRln(P/_{c}P) (Since,_{b}T/_{c}T=_{b}P/_{c}P)_{b}=3/2

nRln(2P/_{0}P)_{0}= 3/2

nRln2From the above observation we conclude that, the change in internal energy Δ

Ein the path_{bc}btocwould be 6p_{0}V_{0}and the change in entropy ΔSin the path_{bc}btocwould be 3/2nRln2.

(c)The change in entropy and internal energy in a cyclic process is zero. Since the process is a complete cycle, therefore the change in entropy and internal energy in going through one complete cycle would be zero.

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Question:3For the Carnot cycle shown in below figure, calculate (a) the heat that enters and (b) the work done on the system.

Concept:In an isothermal process, the heat

Qtransfer takes place at constant temperatureT. The entropy ΔSof the system is defined as,Δ

S=Q/THere

Qis the positive heat flow into the system. So the entropy ΔSof the system will increase.From the equation Δ

S=Q/T, the absorbed heatQwill be,

Q= (ΔS) (T)=

T(S_{f}–S_{i})Here

S_{f}is the final entropy of the system andS_{i}is the initial entropy of he system.For a cyclic path,

Q+W= 0Here

Qis the heat added to the system andWis the work done.

Solution:The above figure shows a Carnot cycle:

(a) We have to calculate the heat

Q_{in}that enters into the system.In the above cycle, the heat only enters along the top path

AB.To obtain the heat

Q_{in}that enters into the system, substitute 400 K forT, 0.6 J/K forS_{f}and 0.1 J/K forS_{i}in the equationQ=T(S_{f}–S_{i}),

Q_{in}=T(S_{f}–S_{i})= (400 K) (0.6 J/K - 0.1 J/K)

= (400 K) (0.5 J/K)

= 200 J

Therefore the heat

Q_{in}that enters into the system would be 200 J.(b) To find the work done, first we have to find out the heat that leaves

Q_{out}from the bottom path of the cycle.In the above cycle, the heat only leaves along the bottom path

CD.

Q_{out}=T(S_{f}–S_{i})= (250 K) (0.1 J/K - 0.6 J/K)

= (250 K) (-0.5 J/K)

= -125 J

Thus the heat that leaves

Q_{out}from the bottom path of the cycle would be -125 J.As,

Q+W= 0 for a cyclic path, so work doneWwill be,

W= -Q= - (

Q_{in}+Q_{out}) (Since,Q=Q_{in}+Q_{out})To obtain the work done

Won the system, substitute 200 J forQ_{in}and -125 J forQ_{out}in the equationW= - (Q_{in}+Q_{out}),

W= - (Q_{in}+Q_{out}) = - (200 J+(-125 J) ) = -75 JFrom the above observation we conclude that, the work done

Won the system would be -75 J.