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FIRST LAW OFTHERMODYNAMICS

Name: Mr. BurnettDate: 24/04/13

Class: 6B

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First Law of thermodynamics

This is an application of the conservation of energy

principle and states that:

- the change in internal energy of a system (U) is

equal to the heat supplied to or removed from it (Q) and

the work done on or by it

U = Q - W

Where,

Q = heat supplied to ortransferred from the

system

W = work done on or by the systemU= change in internal energy of the

system

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Sign Convention

Q is positive when heat is supplied to the

system and negative when heat is transferred

from the system

W is positive if the external work is done bythe system (gas expands) and it is negative if

the work is doneonthe system (gas

compressed). U is positive if the internal energy of the

system increases.

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E.g 1

2500J of heat is added to a system and 1800J

of work is done on the system. What is the

change in internal energy of the system? b)

What would be the internal energy change if2500J of heat is added to the system and

1800J of work is done by the system?

Ans: a) U = 4300J; b) 700J

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Tips to note

The internal energy increases by supplyingheat to the system and/or by doing work on thesystem

For an isolated system, one that does nottransfer heat (Q =0J) and in which nowork is done ( W = 0J) there is no changein internal energy U =0J. Internal energy is

constant. Adiabat ic p rocessesare those that do not

involve any transfer of heat (Q =0J). ThusU= -W. So for such processes the internal

energy of the system is equal to the work

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Work done by a gas

Consider a frictionless piston of surface area,

A, as shown below,

In moving the piston outwards a distance x the gas does somework on the surroundings. The work done (W) here is equal to:

W = F x;

But the force, F, being exerted by the gas is equal to pressure (P) x

Area (A), W = PA. x;

But A .x = Volume (V);

Thus W = p V

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E.g 1

A fixed mass of gas is cooled so that its

volume decreases from 4.0 litres to 2.5 litres at

a constant pressure of 1.0 x 105 Pa. Calculate

the external work done on the gas. (1m3 =1000 litres)

Ans: -1.5 x 102

J

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E.g. 2

A sample of gas is enclosed in a cylinder by a

frictionless piston of area 60cm2. The cylinder

is heated so that 400J of heat energy is

supplied to the gas which then expandsagainst atmospheric pressure and pushes the

piston 20cm along the cylinder. Given that

atmospheric pressure is 1.0 x 105Pa,

calculate:

a) the external work done by the gas (120J)

b) the change in internal energy of the gas

(280J)

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Molar Heat Capacity

Oftentimes we dont know the mass of the gas but we know the number ofmolesthere are in the gas.

Definition: The amount of heat energy required to raise the temperature of1mole of a substance by one kelvin (1K).

There are two (2) forms of the heat formula with molar heat capacities:

For heat transfer at constant pressure: Q = nCpT

For heat transfer at constant volume: Q = nCvT

Where Cp = Molar heat capacity at constant pressure

Cv = Molar heat capacity at constant volume

n = number of moles; Q = heat energy; T= temp change(K)

Units: J/mol/ K or J mol-1 K-1

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Cp vs Cv

The molar heat capacity at constant volume (Cv) is the heat energy

required to produce a unit temperature rise in one mole of a gas

when the volume is kept constant.

The molar heat capacity at constant pressure (Cp) is the heat energy

needed to produce a unit temperature rise in one mole of a gas

when thepressure is kept constant.

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Cp > Cv

Heating a gas at constant

pressure will cause an

increase in its internal

energy and work done

by the gas itself.

Heating a gas at constant

volume will only cause an

increase in the internal

energy of the gas. No work

will be done by or on the

gas.

Constant Pressure,

Cp

Constant Volume,

Cv

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Relation between Cp and Cv

Constant pressure (Cp)

Q = U + W; W =pV

Q= U + pV; Q = nCpT

nCpT = nCvT + pV; pV = nRT where n = 1mol

Cp = Cv + R; R= molar gas constant (8.31 J mol-1 K-1)

Constant volume (Cv)

Q = U + W; W =pV but V = 0m3

Q = U = nCvT

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Tips to Note

The addition or removal of heat does not have

to change the temperature of a body (recall:

latent heat Q = mL).

Cp = Cv + R ; Cp > Cv

The values of Cp and Cv depend on the

universal molar gas constant, R= 8.31 J mol-1

K-1

U = nCvT (always)

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P-V graphs

Identify the lines drawn in p-V graphs above

a) Isotherm ( represents an isothermal process-constant temperature)

b) Isobar(represents an isobaric process

constant pressure)

c) Isochore ( represents an isochoric process

constant volume)

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Work done from p-V graphs

Forisochoric or isovolumetric processes,

there is no change in volume (V =0m3) so

there is no work done, W= 0J.

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Work done during a cycle

The shaded area on thegraph represents the net

work that is done on or by

a gas during 1cycle.

a- b: gas heated up no

work done, W= 0J

(constant volume)

b-c: work done by the gas

at constant pressure (W =

p2V)

c-d: gas cools down at

constant volume (W=0J)d-a: work done on gas at

constant pressure.

(W=p1V)Net work done = Area shaded region

= Area bcfe Area adfe

Or

Net work done = (p2 p1) x (V2 V1)

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PQ

In which steps shown on the curve is work being done onthe gas?

a)A& B

b)C&D

c) A&C

d)B&D

Ans: b

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Tips to note

You should be familiar with the areas of other figures such as the

triangles and trapeziums in order to be competent on these

questions.

1m3 = 1000Litres (L); always convert to S.I units unless told

otherwise

Work done in an isothermal process = W = nRT ln (nat log) (V2)( V1)

isobaric process = W = p V = p (V2 - V1)

isovolumetric/ isochoric process = W = 0J

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E.g.- Work done in a cycle

Five moles of an ideal gas undergo a cycle

consisting of two isobaric processes, ab and cd,

and two isochoric processes, bc and da.

Determine:

a) The work done during the process ab

b) The net work done during the entire cyclec) The temperature at a and b of the gas during

the cycle

d) The heat removed from the gas during the

process c to d (hint: isobaric process) (take

Cv

for gas is = (5/2)R). (take 1atm = 1.01 x

105Pa or N/m2

R = 8.31 J mol-1K-1).

f e

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Solution

a) Given # of moles, n = 5. Now from a to b is an isobaric process (constant pressure),

So the work done from a to b is: W = PV

W = ( 5.0 x 1.01 x 105 )Pa x (62) x 10-3 m3

W = (5.05 X 105 )Pa x (4 x 10-3) m3

W = 2020J or 20.2 x 102 J = Work done over area abfe

b) The net work during the entire cycle: W = (P2 - P1 ) x (V2 - V1)

W = ( 5.051.01) x 105 x (4 x 10-3 m3)

W = 16.16 x 102 J

Or

Net work done in a cycle = Work done for area abfe - Work done for area dcfe= 20.2 x 102(1.01 x 105 )Pa x (4 x 10-3m3 )

= 16.16 x 102J

c) For an ideal gas: pV = nRT; thus the temperatures at points a and b can be

determined:

Ta= paVa / nR = (5.05 x 105

Pa) x ( 2 x 10-3

m3

) / 5 x (8.31)Ta = 24.3K

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Solution cont

Tb = pbVb /nR

Tb = (5.05 x 105Pa) x (6 x 10-3 m3) / 5 x (8.31)

Tb = 72.9K

d) Cv = 5 R

2

Cp = 5 R + R = 7 R

2 2

the heat removed from the isobaric process c - d:

Q = nCpT

Q = 5 x (7 R) x ( 72.9 24.3)

2Q = 5 x (7 x 8.31) x (48.6)

2

Q = 7067.7J

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Try these. Practice makes perfect ;) (y).

Work done in a cycle

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PQ2

The indicator diagram shows an energy cycle for1mole of an ideal gas.

The gas is cooled constant pressure (a to b), heated at constant volume (b

to c) and then returned to its original state (c to a). Calculate:

a) the gas temperature at a, b and c

b) the heat removed from the gas during the process a to b.

c) the heat supplied to the gas during the process b to c, and

d) the net work done in the cycle

(R = 8.31 J mol-1 K-1 ; Cv = (5/2)R

a

c

b

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PQ- Net work done in a cycle

A fixed mass of gas is taken through the

closed cycle ABCD as shown in the diagram.

Calculate the work done by the gas during this

cycle of events.

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PQ2- Net work done in a cycle

The accompanying pressure-volume graph shows an engine cycle with one

isothermal process, B-C. The working substance is an ideal gas. At point

B, the pressure is 3.20x105 Pa and the volume is 0.0480 m3. At point C, the

pressure is 2.10x105Pa and the volume is 0.0730 m3. At point A the volume

is 0.0230 m3. What is,

a) The gas temperature at A, B, C and D?

b) The net work done by the gas during one cycle?

c) The heat energy removed or supplied during the processes a to b, b to c

and c to d? (R = 8.31 J/mol/K; Cp = 29 J mol-1 K-1 ).