Heat capacity = C = amount of heat the object receives for a 1K temperature rise = does not depend on the mass = unit J K–1

C Heat it can take

For 1 K temperature rise

Specific Heat capacity = c = amount of heat required to raise the temperature of 1 kg of object by 1K = depend on the mass = unit J kg–1 K–1

c = C m

4200Water

2000Ice

138Mercury

470Iron

390Copper

910Aluminium

Specific heat capacity (J kg–1 K–1)

Subtance

C Heat it can take

For 1 K temperature rise

Molar Heat capacity = Cm

= amount of heat required to raise the temperature of 1 mole of subtance by 1K = depend on the n = unit J mol–1 K–1

Cm = mrc 1000

Specific Heat capacity = c = amount of heat required to raise the temperature of 1 kg of object by 1K = depend on the mass = unit J kg–1 K–1

Relative molecular/atomic mass

75.4Water

36.5Ice

27.7Mercury

26.3Iron

24.8Copper

24.6Aluminium

Molar heat capacity (J mol–1 K–1)

Subtance

Q = C∆θ

Q = mc∆θ

Q = nCm∆θ

Amount of heat required to raise the temperature

F = pA

dW = F dy

dW = pA dy

= p dV

∫ dW = ∫ p dV

W = ∫ p dVV2

V1

dy

Container and piston

A

heating

Pressure = p F

dV

W = ∫ p dVV2

V1

Boyle’s lawConstant pressure

P ∝ 1V

P = k V

W = ∫ dVV2

V1

kV

= k ∫ dVV2

V1

1V

= k [ ln V ] V2

V1

= k [ln V2 – ln V1] = k ln

expansion

compression

V not change

V2 > V1

V2 < V1

V2 = V1

ln

V 2V1

V 2V1

= +ve

W = +ve

ln V 2V1

= –ve

W = –ve

ln = ln 1 = 0

W = 0

V 2V1

W = p ∫ dVV2

V1

= p[ V ]V2

V1

= p[V2 – V1]

= p∆V

W = ∫ p dVV2

V1

Constant pressure

= k ln

expansion

compression

V not change

V2 > V1

V2 < V1

V2 = V1

ln

V 2V1

V 2V1

= +ve

W = +ve

ln V 2V1

= –ve

W = –ve

ln = ln 1 = 0

W = 0

V 2V1

W = p ∫ dVV2

V1

= p[ V ]V2

V1

= p[V2 – V1]

= p∆V

W

expansion

compression

V2 > V1

W = +ve

V2 < V1

W = –ve

P

0 V

expanding

V1 V2

P

0 V

compressed

V2 V1

Work done by gas

Work done on gas

∆Q = ∆U + W

Reflects the principle of conservation of energy

Heat energy

supplied

Increase in

internal energy

Work done

by the gas

∆Q = ∆U – W

Work done

on the gas

+veHeat

supplied

–ve Heat loss

+veIncrease

ininternalenergy

–ve Decrease

ininternal energy

Example :

820 cm3 1000 cm3

Heat = 220 J

T V

P = 2 x 105 Pa

a) Work done by the gas ?b) Change in internal energy ?

Solution :

P constant W = p ∫ dVV2

V1

= p ∆V

= (2 x 105)[(1000 x (10–2)3) – 820 x (10–2)3]

= + 36 J

a)

b) ∆Q = ∆U + W

∆U = ∆Q – W = 220 – 36

= + 184 J

+ve means internal energy increases

Molecules have

KE

PE

Always in motion

Forces of attraction between molecules

∆U∆Q = ∆U + W

∆Q

heating

KE

∆U

∆Q = ∆U + W

∆U = ∆Q – W

Case where∆U

When work done on gas, W = –ve

∆U = ∆Q – ( – W)

= ∆Q + W

∆U is +ve

∆U∆Q = ∆U + W

∆Q

heating

KE

∆U

∆Q = ∆U + W

∆U = ∆Q – W

Case where∆U

When work done on gas, W = –ve

∆U = ∆Q – ( – W)

= ∆Q + W

∆U is +ve

Conclusion :

∆U By heating

Performing mechanical work on gas

For ideal gas :

No force of attraction the separation is large

No PE involve,Only KE involves

KE ∝ T

FromMolecular KE

= 3/2 KT

∆U ∝ T

T constant∆U = 0

∆Q = nCp,m ∆T

∆Q = nCV,m ∆T

Heat supplied to increase the temperature of n moles of gas by 1K at constant pressure :

Heat supplied to increase the temperature of n moles of gas by 1K at constant volume :

Molar heat capacity at constant pressure

Molar heat capacity at constant volume

Change of state of one mole of an ideal gas

Case 1 Case 2

pVT

p + ∆pVT + ∆T

pVT

pV + ∆VT + ∆T

Work done by a gas :W = p∆V

= p(0)

= 0

∆Q = CV,m∆T∆Q = ∆U + W

CV,m∆T = ∆U + 0

∆U = CV,m∆T

∆Q = ∆U + W∆Q = Cp,m∆T

Work done by a gas :W = p∆V

pVm = RT

p(Vm + ∆V) = R(T + ∆T)

--(1)

pVm + p∆V) = RT + R∆T --(2)

(2) – (1) : p∆V = R∆T

= R∆T

Cp,m∆T = CV,m∆T + R∆T

Cp,m = CV,m + R

Cp,m – CV,m = R

∆U = CV,m∆T

From case 1 :

Cp,m – CV,m = R

Cp.m > CV,m

0

p

V Vm

p

p + ∆pCase 1 (isochoric process)

Vm + ∆V

Case 2 (isobaric process)

γγ =

Cp,m

CV,mRatio of principal molar heat capacities

Cp,m and CV,m depend on the degrees of freedom

One mole of ideal gas has internal energy : U = RTf2

∆Q = ∆U + W

CV,m∆T = ∆U∆Q = CV,m∆T W = p∆V = p(0) = 0

CV,m = ∆U

∆T CV,m = dU

dT

As ∆T 0

= Rf2

Cp,m – CV,m = R Cp,m = R + Rf2

= Rf + 2

2

It is shown that Cp,m and CV,m depend on degrees of freedom

γ =Cp,m

CV,m

=

= f + 2 f

For monoatomic : f = 3

γ = 3+2 3

= 1.67

For diatomic : f = 5

γ = 5+2 5

= 1.4

For polyatomic : f = 6

γ = 6+2 6

= 1.33

Constant temperature

Internal energy constant

∆U = 0

pV = constant

Boyle’s law

p

0 V

T1

T2

T3

T1 < T2 < T3

isotherm

Isothermal compression Isothermal expansion

Work done on gas

Heat escapes

T constant

Work done by gas

Heat enters

T constant

pV = nRT

p = nRT V

W = ∫ p dVVf

Vi

= ∫ dVVf

Vi

nRT V

= nRT ∫ dVVf

Vi

1V

= nRT [ ln V ]Vf

Vi

= nRT [ln Vf – ln Vi]

= nRT ln V fVi

0 Vi Vf V

p

pi

pf

Isothermal expansion

WorkDone

ByIdealgas

W = +ve Vf > Vi

piVi = pfVf

= nRT ln V fVi

W

pV = nRT= pfVf lnV fVi

piVi lnV fVi

First law of thermodynamic : ∆Q = ∆U + W

∆Q = W ∆U = 0

Work done on the gas

pV = nRT

p = nRT V

W = ∫ p dVVf

Vi

= ∫ dVVf

Vi

nRT V

= nRT ∫ dVVf

Vi

1V

= nRT [ ln V ]Vf

Vi

= nRT [ln Vf – ln Vi]

= nRT ln V fVi

0 Vf Vi V

p

pf

pi

Isothermal compression

WorkDone

onIdealgas

W = –ve Vf < Vi