Heat capacity is the amount of heat required to raise the temperature of an object by 1K. Specific heat capacity is the heat required to raise the temperature of 1 kg of a substance by 1K. Molar heat capacity is the heat required to raise the temperature of 1 mole of a substance by 1K. The work done by a gas during expansion or compression depends on the change in volume and pressure. For an ideal gas, the molar heat capacities Cp,m and CV,m depend on the degrees of freedom and Cp,m is always greater than CV,m.

3. 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
c =C
m
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

4. Subtance Specific heat capacity
(J kg–1 K–1)
Aluminium 910
Copper 390
Iron 470
Mercury 138
Ice 2000
Water 4200

5. 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 Relative
molecular/atomic mass
C Heat it can take
Cm = mrc
1000
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

6. Subtance Molar heat capacity
(J mol–1 K–1)
Aluminium 24.6
Copper 24.8
Iron 26.3
Mercury 27.7
Ice 36.5
Water 75.4

7. Q = C∆θ
Amount of heat
required to raise Q = mc∆θ
the temperature
Q = nCm∆θ

8. F = pA
A dW = F dy dV
dW = pA dy
Pressure = p F
= p dV
dy ∫ dW = ∫ p dV
heating
V2
Container and piston W= ∫ p dV
V1

9. V2 expansion
W= ∫ p dV V2 > V 1
V1
ln V2 = +ve
Boyle’s law V1
Constant W = +ve
P∝ 1
V pressure
P= k V2
compression
W=p∫ dV
V V2 < V 1
V1 V2
V2 k
W= ∫ dV = p[ V ] ln V2 = –ve
V V1 V1
V1
V2 1 = p[V2 – V1] W = –ve
= k ∫ V
dV
V1 = p∆V V not change
V2 V2 = V 1
= k [ ln V ]
V1 ln V2 = ln 1 = 0
V2 V1
= k [ln V2 – ln V1] = k ln
V1 W=0

10. V2 expansion
W= ∫ p dV V2 > V 1
expansion V1
V2 > V 1 ln V2 = +ve
V1
W = +ve Constant W = +ve
pressure
V2
compression
W=p∫ dV
V1 V2 V2 < V 1
= p[ V ] ln V2 = –ve
V1 V1
compression = p[V2 – V1] W = –ve
V2 < V 1 = p∆V V not change
W = –ve V2 = V 1
ln V2 = ln 1 = 0
V2 V1
W = k ln
V1 W=0

11. P
Work done by gas
expanding
0 V1 V2 V
P
Work done on gas
compressed
0 V2 V1 V

12. Reflects the principle of conservation of energy
∆Q = ∆U + W ∆Q = ∆U – W
Heat Increase Work Work
energy in done done
supplied internal by the on the
energy gas gas

13. +ve –ve –ve
Heat Heat
+ve
Increase Decrease
supplied loss in
in
internal internal
energy energy

14. Example :
T V
820 cm3 1000 cm3
P = 2 x 105 Pa
Heat = 220 J a) Work done by the gas ?
b) Change in internal energy ?
Solution :
V2
a) P constant W=p∫ dV = p ∆V
V1
= (2 x 105)[(1000 x (10–2)3) – 820 x (10–2)3]
= + 36 J

15. b) ∆Q = ∆U + W
+ve means
∆U = ∆Q – W = 220 – 36 internal energy
increases
= + 184 J

17. KE Always in motion
Molecules have
PE Forces of attraction
between molecules
∆Q = ∆U + W ∆Q = ∆U + W
∆U
∆U = ∆Q – W
heating
When work done on gas,
∆Q W = –ve
Case ∆U = ∆Q – ( – W)
KE
where = ∆Q + W
∆U ∆U 
∆U is +ve

18. Conclusion :
By heating
∆U 
Performing mechanical work on gas
∆Q = ∆U + W ∆Q = ∆U + W
∆U
∆U = ∆Q – W
heating
When work done on gas,
∆Q W = –ve
Case ∆U = ∆Q – ( – W)
KE
where = ∆Q + W
∆U ∆U 
∆U is +ve

19. For ideal gas :
No force of attraction the separation is large
No PE involve,
Only KE involves
From
Molecular KE T constant
KE ∝ T
= 3/2 KT ∆U = 0
∆U ∝ T

21. Heat supplied to increase the temperature of n moles of gas
by 1K at constant pressure :
∆Q = nCp,m ∆T
Molar heat capacity at constant pressure
Heat supplied to increase the temperature of n moles of gas
by 1K at constant volume :
∆Q = nCV,m ∆T
Molar heat capacity at constant volume

24. Change of state of one mole of an ideal gas
Case 1 Case 2
p p + ∆p p p
V V V V + ∆V
T T + ∆T T T + ∆T

25. Work done by a gas :
W = p∆V
= p(0)
=0
∆Q = ∆U + W
∆Q = CV,m∆T CV,m∆T = ∆U + 0
∆U = CV,m∆T

26. Work done by a gas :
W = p∆V = R∆T
pVm = RT --(1)
p(Vm + ∆V) = R(T + ∆T)
pVm + p∆V) = RT + R∆T --(2)
(2) – (1) : p∆V = R∆T
∆Q = Cp,m∆T
∆Q = ∆U + W
From case 1 : Cp,m∆T = CV,m∆T + R∆T
∆U = CV,m∆T Cp,m = CV,m + R
Cp,m – CV,m = R

27. Cp,m – CV,m = R
Cp.m > CV,m

28. p
Case 1 (isochoric process)
p + ∆p Case 2 (isobaric process)
p
V
0 Vm Vm + ∆V

29. γ
Cp,m
γ = Ratio of principal molar heat capacities
CV,m
Cp,m and CV,m depend on the degrees of freedom

30. f
One mole of ideal gas has internal energy : U= RT
2
∆Q = ∆U + W
∆Q = CV,m∆T CV,m∆T = ∆U W = p∆V = p(0) = 0
CV,m = ∆U
∆T
As ∆T  0 CV,m = dU
dT
f It is shown
= R
2 that Cp,m
f and CV,m
Cp,m – CV,m = R Cp,m = R+ R
2
depend on
f+2 degrees of
= R
2 freedom

31. Cp,m
γ = =
CV,m
= f +2
f
For monoatomic : f = 3
γ = 3+2 = 1.67
3
For polyatomic : f = 6
For diatomic : f = 5 γ = 6+2 = 1.33
γ = 5+2 = 1.4 6
5

33. Constant temperature pV = constant
Boyle’s law
p
Internal energy constant T1 < T2 < T3
∆U = 0 isotherm
T3
T2
T1
0 V

34. Isothermal compression Isothermal expansion
Work Work
done on done by
gas gas
T constant T constant
Heat escapes Heat enters

35. Vf
pV = nRT W=∫ p dV
Vi p
Vf
p = nRT =∫ nRT dV
Isothermal
V Vi V pi
Vf 1
expansion
= nRT ∫ V dV pf
Vi
Work Vf
Done = nRT [ ln V ] 0 Vi Vf V
By Vi
Ideal
gas = nRT [ln Vf – ln Vi]
= nRT ln Vf W = +ve  Vf > Vi
Vi

36. W = nRT ln Vf
Vi
piVi = pfVf piVi ln Vf = pfVf ln Vf pV = nRT
Vi Vi
First law of thermodynamic : ∆Q = ∆U + W
∆Q = W ∆U = 0

37. Work done on the gas

38. Vf
pV = nRT W=∫ p dV
Vi p
Vf
p = nRT =∫ nRT dV
Isothermal
V Vi V pf
Vf 1
compression
= nRT ∫ V dV pi
Vi
Work Vf
Done = nRT [ ln V ] 0 Vf Vi V
on Vi
Ideal
gas = nRT [ln Vf – ln Vi]
= nRT ln Vf W = –ve  Vf < Vi
Vi