thermodynamics of gases2
TRANSCRIPT
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Not allowing heat to enter or leave the gas
∆Q = 0 ∆Q = ∆U + W
0 = ∆U + W
∆U = –W
Thermodynamic first law
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∆U = –W
Work done by the gas
∆U = –W
W = +ve
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∆U = –W
Work done by the gas
∆U = –W
W = +veWork done on the gas
∆U = –(–W) = +W
W = –ve
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T1V1γ – 1 = T2V2
γ – 1
p11 – γ T1
γ = p21 – γ T2
γ
back
back
back
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∆U
∆T
∆Q = 0
∆U = CV,m∆T
W = p∆V
∆Q = ∆U + W
0 = CV,m∆T + p∆V
Thermodynamic first law
p∆V = –CV,m∆T ---------(1)back
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Initially:pV = RT
After compression:(p + ∆p)(V + ∆V) = R(T + ∆T)
For one mole of an ideal gas
pV + p∆V + V∆p + ∆p∆V = RT + R∆T)
substitute
RT + p∆V + V∆p + ∆p∆V = RT + R∆T)Assumed zero
p∆V + V∆p = R∆T
V∆p = R∆T – p∆V Substitute (1) into here∆pV = R∆T + CV,m∆T
∆pV = (R + CV,m )∆T Cp,m – CV,m = R
= Cp,m ∆T ---------(2)
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∆pV = Cp,m ∆T ---------(2)
(2)(1)
:
∆p p
= –Cp,m
CV,m
∆VV
– γ ∆VV
=∆p p
γ =Cp,m
CV,m
---------(3)
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– γ dVV
=dp p
As ∆p 0 and∆V 0
– γ dVV
=dp p∫ ∫1p
dp =∫ – γ ∫ 1V
dV
ln p = – γ ln V + m
ln p = ln V–γ + m Use log rule
p = e ln V–γ + m Use log rule
p = e eln V–γ m
p = e (constant)ln V–γ
p = V–γ (constant)Use log rule
pVγ = constant ----------(4)
p1V1γ = p2V2
γ
To adiabatic equation
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pVγ = constant
p = RT V
RTV Vγ = constant
RTVγ – 1 = constant
TVγ – 1 = constant R
TVγ – 1 = constant ----------(5)
T1V1γ – 1 = T2V2
γ – 1
Because R is also a constant
To adiabatic equation
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pVγ = constant
V = RT pp
RTp
γ= constant
p1 – γ
Rγ
Tγ
= constant
p1 – γ
Tγ
=constant
Rγ
p1 – γ
Tγ
= constant
Because R γ is also a constant
p11
– γ T1 = p2
1 – γ T2
γ γ
To adiabatic equation
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0
P
V
T1
T2
Adiabatic expansionpi
Vi
pf
Vf
W = ∫ p dVVf
Vi
From pVγ = constant= k
p = kV–γ = ∫ kV–γ dV
Vf
Vi
= k ∫ V–γ dV
Vf
Vi
= kVf
Vi
V–γ+1
–γ+1
V–γ+1=k
–γ+1
Vf
Vi
=k
–γ+1 Vf–γ+1 Vi
–γ+1–
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----(1)
=1
1 – γ kVf–γ+1 kVi
–γ+1– pVγ = constant = k
piViγ = pfVf
γ = k=
1
1 – γ ( ) Vf–γ+1 ( ) Vi
–γ+1–pfVfγ
piViγ
=1
1 – γ –pfVfpiVi
=1
γ – 1 –piVi pfVf ----(2)
pV = nRT=
1
γ – 1 –nRT1 nRT2
=nR
γ – 1 –T1 T2----------(3)
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=nR
γ – 1 –T1 T2W ∆T = T2 – T1
γ = f + 2 f =
nR
– 1
( )–∆Tf + 2 f
=nR
f + 2 f
– ff
(–∆T)
=nR2f
(–∆T)
=nfR 2
– ∆T
= –∆U
W = –∆U
Work done by the gas
Work done on the gas , W = –ve
W = +∆U
Means a rise in its internal energy
T W = –∆UMeans a reduction in its internal energy
T
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Is a process which can be made to retrace its path from one equillibrium state to another equillibrium state through small changes at every step.
Isothermal expansion Isothermal compression
p
0 V
Small step at every instant
p
0 V
Small step at every instant
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Isothermal expansion Isothermal compression
p
0 V
Small step at every instant
p
0 V
Small step at every instant
The wall of the container must be as thin as possible to allow heat transfer.
The piston must be light and frictionless.carried out very slowly through small steps
For a reversible process to occur
in practicce
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For a reversible process to occur
in practicce
Thick insulator so that heat transfer cannot occur
Piston must be light and frictionless
Must be carried out very quickly through small steps
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Example 1 :
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Example 2 :
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Example 3 :
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