1. Dr. S. B Maulage
Dept of Chemistry

2. Concerned with the study of transformation of energy: Heat  work
THERMODYNAMICS.
Concerned with the study of transformation of energy:
Heat  work

3. CONSERVATION OF ENERGY – states that:
Energy can neither be created nor destroyed in chemical reactions. It can only be converted from one form to the other.
UNIVERSE
System – part of world have special interest in…
Surroundings – where we make our observations

4. → → ↔ matter ↔ energy ↔ energy not matter matter × Energy × Example:
→ →
Open system
Closed system
Isolated system

5. If system is thermally isolated called Adiabatic system eg: water in vacuum flask.

6. HCl(aq) + NaOH(aq) → NaCl(aq) + H2O(l) - heat given off.
WORK and HEAT
Work – transfer of energy to change height of the weight in surrounding eg: work to run a piston by a gas.
Heat – transfer of energy is a result of temperature difference between system and surrounding eg:
HCl(aq) + NaOH(aq) → NaCl(aq) + H2O(l) - heat given off.
If heat released to surroundings – exothermic.
If heat absorbed by surroundings – endothermic.

7. Example: Gasoline, 2, 2, 4 trimethylpentane
CH3C(CH3)2CH2CH(CH3)CH3 + 25/2 O2 → 8CO2(g) +H2O(l)
5401 kJ of heat is released (exothermic)
Where does heat come from?
From internal energy, U of gasoline. Can represent chemical reaction:
Uinitial = Ufinal + energy that leaves system (exothermic) Or
Ui = Uf – energy that enters system (endothermic)

8. Hence, FIRST LAW of THERMODYNAMICS (applied to a closed system)
The change in internal energy of a closed system is the energy that enters or leaves the system through boundaries as heat or work. i.e.
∆U = q +w
∆U = Uf – Ui
q – heat applied to system
W – work done on system
When energy leaves the system, ∆U = -ve i.e. decrease internal energy
When energy enter the system, ∆U = +ve i.e. added to internal energy

9. Different types of energies:
Kinetic energy = ½ mv2 (chemical reaction) kinetic energy
(KE)  k T (thermal energy) where k = Boltzmann constant
Potential energy (PE) = mgh – energy stored in bonds
Now, U = KE + PE

10. 3. Work (W) w = force × distance moved in direction of force i. e
3. Work (W) w = force × distance moved in direction of force i.e. w = mg × h = kg × m s-2 × m = kg m2 s-2 (m) (g) (h) 1 kg m2 s-2 = 1 Joule - Consider work – work against an opposing force, eg: external pressure, pex. Consider a piston

11. Piston

12. w = distance × opposing force w = h × (pex × A) = pex × hA
Work done on system = pex × ∆V
∆V – change in volume (Vf – Vi)
Work done by system = -pex × ∆V
Since U is decreased

13. Example:
C3H8(g) + 5 O2(g) → 3CO2(g) + 4H2O(l) at 298 K  1 atm (1 atm = Pa), kJ = q What is the work done by the system? For an ideal gas; pV = nRT (p = pex) n – no. of moles R – gas constant T = temperature V – volume p = pressure

14. V= nRT/p or Vi = niRT/pex
6 moles of gas:
Vi = (6 × × 298)/ = m3
3 moles of gas:
Vf = (3 × × 298)/ = m3
work done = -pex × (Vf – Vi)
= ( – ) = J

15. NB: work done = - pex (nfRT/pex – niRT/pex) = (nf – ni) RT Work done = -∆ngasRT i.e. work done = - (3 – 6) × × 298 = J Can also calculate ∆U ∆U = q +w q = kJ w = J = 7.43 kJ  ∆U = = kJ

16. NB: qp  ∆U why? Only equal if no work is done i.e. ∆V = 0
i.e. qv = ∆U

17. Example: energy diagram

18. Since work done by system = pex∆V
System at equilibrium when pex = pint (mechanical equilibrium)
Change either pressure to get reversible work i.e.
pex > pint or pint > pex at constant temperature by an infinitesimal change in either parameter

19. For an infinitesimal change in volume, dV Work done on system = pdV
For ideal gas, pV = nRT
p = nRT/ V
 work =  p dV = nRT dV/ V
= nRT ln (Vf/Vi) because dx/x = ln x
Work done by system= -nRT ln(Vf/Vi)

20. Work done along adiabatic path
ie q = 0 , no heat enters or leaves the system.
Since U = q + w and q =0
U = w
When a gas expands adiabatically, it cools.
Can show that: pVγ = constant, where ( Cp/Cv =γ )
and: piViγ = pfVfγ and since:
-p dV = Cv dT