lecture powerpoint

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