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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 ).