UNIVERSITY OF NAIROBI Lecture 2

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UNIVERSITY OF NAIROBI

Lecture 2

1ST Law of Thermodynamics

1ST Law

Thermodynamic Processes

Heat capacities


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Objectives

Explain 1ST Law, its significance and limitations

Explain various thermodynamic processes.

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Introduction

A Donkey is an engine: It eats (= heat input) to get

energy in order to work -Concept of 1ST Law


RECAL:

Energy may be converted from one form to another e.g., heat to work but TOTAL ENERGY MUST BE CONSERVED

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1ST Law of Thermodynamics

Defines relationship between heat (dQ), work (dW) & internal energy (dU) of a system by mandating conservation of energy.


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dQ = dU + dW

Heat flow into/out of system

Change in internal energy

Work done by (or on) system

1ST Law

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1ST Law Simplified

“We eat in order to work’

Or

“No animal works continously without eating or wearing out”

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dQ = dU + dW


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For mechanical device e.g., gas in cylinder

dW = Fdx = PA dx dW = P(Adx) = PdV

dQ = dU + PdV ………………….(2)

Gas dQ

dx Area A

Movable piston P

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Internal Energy of Ideal Gas

From Kinetic Theory of gases, internal

energy (U) of ideal gas with f degrees of

freedom is:

f = translational & rotational degrees

of freedom. Vibrational ones are

negligible

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f 3 (monatomic), 5 (diatomic) etc

)1.......(..........22 TNKRTUB

ff


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Note

U is a state function i.e., it does not depend on the path taken/process

dUabc = dUadc S.T

U depends only on T

For reversible process,

dU = 0

P

V

T a

B

b

c

d

C

dU 0

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dQ & dW are path dependent & are not state functions S.T dWabc dWadc

C C

dQdW 0&0


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Significance of 1ST Law

It provides means of determining dU

Limitations of 1ST Law Does not define efficiency of systems i.e.,

how much dQ is converted into dW

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Perpetual Machines of 1st kind

A machine that works continously without energy input (dQ) or change in (dU) does not exist.


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Worked Example

A device working reversibly takes in 1000 J of heat, ejects 200 J of heat to a cold reservoir and produces 800 J of work.

Does this device violate 1ST Law ?

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Soln

(i) 1ST Law mandates conservation of Energy

Since system is reversible dU = 0

dW = dQ

W (800) = Qh (1000) - Qc (200)=800J

device doesn't violate 1ST Law since no energy is created nor destroyed.


Thermodynamics and Laws of nature Rule of Nature: We eat to live

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Laws of Nature Laws of Thermodynamics

1. We eat when hungry

Zeroth Law: Heat flows from hot to Cold

2. We eat in order to work

1st Law: dQ = dU + dW

3. (i) If you eat, you must shit

(ii) You live once

2nd Law: (i) Efficiency of systems <

100% (ii) Nature is irreversible,

dS 0

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Lecture Evaluation

State 1ST Law giving its significance and limitations

Explain perpetual machines of 1ST Kind

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Thermodynamic processes

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Process = passage of a system from an initial to a final state of equilibrium as a result of transfers of heat, work or Internal Energy

An ideal gas can undergo 4 main Processes:-

isothermal (heat transfer at const temp)

isobaric (const P)

isochoric (const V) and

adiabatic (no heat input)


Other Processes

o Reversible process → the system is

close to equilibrium at all times. NB.

there are no truly reversible processes

in nature.

o Cyclic process → the final and initial

state are the same.

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These processes can be represented on P-V diagram

Isothermal (dT=0)

V

P

a

Adiabatic (dQ=0)

c Isobaric

(dP = 0)

Isochoric

(dV = 0)

d b


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Applications of 1ST Law

Used to find work done in various thermodynamic processes for purposes of designing a (heat engine) with max possible efficiency

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(a) Isothermal process (dT = 0)

Work done = area under P-V curve i.e.,

but

Also since

2

1

21

V

V

PdVW

V

P

PV = c

V2 V1

W 2

1

isothermal V

nRTP

1

2ln

2

1

V

VnRT

V

dVnRTW

V

V

2211 VPVP 2

1lnP

PnRTW


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(b) Isochoric Process (dV = 0)

heat input only increases dU

V

P

V1,2

T1

T2

1

2

gasmonoatomicforRdTdUdQ2

3

0

2

1

21

PdVW

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(c) Isobaric process (dP = 0)

1ST Law becomes:

dQ = dU + PdV = dH

V

P

V1

1

2

V2

Where dH = Entalpy = heat involved in chemical

reaction (at const P)

012

2

1

21

VVPPdVW


Worked Example

An ideal gas is taken through cyclic process ABCA. Determine

(i) Net heat transferred to system during one complete cycle

(ii) Net heat input for reversed cycle ACBA.

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V m3

P

B

6 8 10

C A

6

2

8

4

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Soln

(i) From first law, dQ = dU + dW

But dW = PdV = area under P-V dU = 0 in cyclic process.

dQ = PdV = ½(8-2)(4) = 12 Kj (ii) dQ is the same as in (i)

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Activity 2

An Ideal gas is taken from state a to c either through abc or adc. In processes ab and bc, 600 J and 200J are given to the system respectively. Calculate

(i) dU along ab (ii) dU along abc (iii) Total heat added in adc

2

3

8

P104 Pa

b c

a d

5 V10-3 m3

200J

600J

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Solution

From 1ST Law (dU = dQ – dW)

(i) Along ab, dW = 0: dU = dQ = 600 J (ii) Total dQ along a b c = (600 + 200) = 800 J

dW along ab = 0 dW along bc = PdV = (8 104Pa)(3 10-3m3) = 240 J.

Total dW along a b c = 240 J dU along a b c = dQ - dW = (800 - 240) J = 560 J


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(iii) Along a d c, dU = 560 J

(NB. dU is a state function S.T total dU along a b c = dU along a d c.)

dW along a d c

= P1(V2 - V1) = 90 J

NB. along d c, dW = 0

Total dQ in adc

= dU + dW

= (560 + 90)J = 650J. 2

3

8

P

b c

a d

5 V

200J

600J

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(d) Adiabatic process (dQ = 0)

No heat enters or leaves the system BUT gas expands doing work (dW) as dU decreases with decrease in T (path 1-2)

V

P

V2

2

1

V1

Adiabatic

Isothermal, PV =c


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But

To obtain dW, we first need to obtain Eqn for adiabatic process by substituting for dT in terms P and V in eqn (3).

From the ideal gas law (PV = nRT), differentiating we have

RdTVdPPdV

)2.....(..........23 RdTRdTdU

)1.......(10 LawofFormdUPdV st

)3....(....................0 RdTPdV

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Substituting for dT in (3), gives

Integrating we have

VdPPdV

VdPPdVPdV

1

0

cPV

P

dP

V

dV

lnln1

1

R

VdPPdVdT


Adiabatic curve has steeper slope ( times) than isothermal curve

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)5........(

2

51ln

AdiabaticforEqncPV

wherecPV

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Activity

Show also that for adiabatic process

cTPii

cTVi

1

)(

)(


Soln

From

Replacing for P from ideal gas Eqn (PV = nRT)

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cPV

cTV

VTVT

V

V

VnRT

VnRT

1

1

22

1

11

1

2

22

11

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Work done in adiabatic Process

W = area under curve

V1 V2 V m-3

P 1

Adiabatic, PV = c P1

P2 2

W

cPV


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since

But Replacing C we get

2

1

2

1

V

V

V

VV

dVCPdVW

V

CP

.2211 CVPVP

1

2211

VPVPW

1

1

1

21

1 V

C

V

CW

1

11

2

22

1

1

V

VP

V

VPW

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Alternatively

Substituting fot T1 and T2 from PV = nRT, we get

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2

1

2

1

1.23

T

T

V

V

nfordTRdUPdVdW

2123

1223 TTRTTR

1

22112211

23

VPVP

R

VP

R

VPRdW


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Activity

Show that W can also be given by

(i)

21

1TT

RW

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22 2

1 1 1

11 11 2 1 1

1 1 1 1

2 1

1( , )

1

1 1 1

1

VV V

V V V

PVW P V T dV dV PV V

V

PVV V

constVPPV

11(i) From

Soln


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Examples of Adiabatic processes

Adiabatic processes take place rapidly e.g.,

(i) Air compressor

Air let out from an air compressor feels cold

(ii) Opening a bottle of carbonated beverage

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Activity

A gas is rapidly compressed from initial

pressure is 105 Pa and temperature of

220C (= 295 K) to a volume that is a

quarter of its original volume (e.g.,

pumping bike’s tire). Find the final

temperature?

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Soln

From

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cVTVT 1

22

1

11

KV

VTT 514

1

2

1

12

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Activity 2.4

During the ascent of a meteorological helium-gas balloon, its volume increases from Vi = 1 m3 to Vf = 1.8 m3, and pressure inside balloon decreases linearly from 1 bar (=105 N/m2) to 0.5 bar.

(a) If initial T is 300K, find the final T ?

(b) How much work is done by the gas?

(c) How much “heat” does the gas absorb?

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Lecture -Evaluation

Explain the following Processes –Isothermal, Isobaric, Isochoric, Adiabatic

State Eqns of state for Isothermal & Adiabatic & give expressions for work done in these processes

List some applications of Adiabatic process

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UNIVERSITY OF NAIROBI 2.3

Heat Capacities of Ideal Gas


Heat Capacities of Ideal Gas

Heat absorbed by a system (dQ) is measure by its capacity to absorb heat i.e., its heat capacity (C).

The heat capacity of a system is the amount of heat energy required to produce a unit temperature rise in that system

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dT

dQC

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NB. Heat can be added to a system either at const Pressure (P) or constant Volume (V)

For a constant volume process, the specific heat capacity (Cv) is:- (NB dW =0)

)1(..........2

3R

dT

dU

dT

dQC

VV

V


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If dQ is added at constant pressure, then specific heat capacity at const P (CP) is

But from (PV = nRT )

Cp - Cv = R Mayer’s relation

PP

PdT

PdVdU

dT

dQC

)2.........(RCdT

nRdT

dT

dUC VP

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Ratio of heat Capacities

)3..(..............................67.1 V

P

C

C The Gamma relation


Summary

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dT

PdVdU

dT

QC

RT

UC

V

V2

3

RCT

HC

V

P

P

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Lecture -Evaluation

Explain Cp and Cv

Why is Cp > Cv always ?


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UNIVERSITY OF NAIROBI Lecture 2