thermodynamics of gases2

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Page 1: Thermodynamics of gases2
Page 2: Thermodynamics of gases2

Not allowing heat to enter or leave the gas

∆Q = 0 ∆Q = ∆U + W

0 = ∆U + W

∆U = –W

Thermodynamic first law

Page 3: Thermodynamics of gases2

∆U = –W

Work done by the gas

∆U = –W

W = +ve

Page 4: Thermodynamics of gases2

∆U = –W

Work done by the gas

∆U = –W

W = +veWork done on the gas

∆U = –(–W) = +W

W = –ve

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Page 7: Thermodynamics of gases2

T1V1γ – 1 = T2V2

γ – 1

p11 – γ T1

γ = p21 – γ T2

γ

back

back

back

Page 8: Thermodynamics of gases2

∆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

Page 9: Thermodynamics of gases2

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)

Page 10: Thermodynamics of gases2

∆pV = Cp,m ∆T ---------(2)

(2)(1)

:

∆p p

= –Cp,m

CV,m

∆VV

– γ ∆VV

=∆p p

γ =Cp,m

CV,m

---------(3)

Page 11: Thermodynamics of gases2

– 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

Page 12: Thermodynamics of gases2

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

Page 13: Thermodynamics of gases2

pVγ = constant

V = RT pp

RTp

γ= constant

p1 – γ

= constant

p1 – γ

=constant

p1 – γ

= constant

Because R γ is also a constant

p11

– γ T1 = p2

1 – γ T2

γ γ

To adiabatic equation

Page 14: Thermodynamics of gases2

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–

Page 15: Thermodynamics of gases2

----(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)

Page 16: Thermodynamics of gases2

=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

Page 17: Thermodynamics of gases2

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

Page 18: Thermodynamics of gases2

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

Page 19: Thermodynamics of gases2

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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