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

∆U = –W

Work done by the gas

∆U = –W

W = +ve

∆U = –W

Work done by the gas

∆U = –W

W = +veWork done on the gas

∆U = –(–W) = +W

W = –ve

T1V1γ – 1 = T2V2

γ – 1

p11 – γ T1

γ = p21 – γ T2

γ

back

back

back

∆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

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)

∆pV = Cp,m ∆T ---------(2)

(2)(1)

:

∆p p

= –Cp,m

CV,m

∆VV

– γ ∆VV

=∆p p

γ =Cp,m

CV,m

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

– 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

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

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

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–

----(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)

=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

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

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

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

Example 1 :

Example 2 :

Example 3 :