Dr. S. B Maulage Dept of Chemistry

Dr. S. B Maulage Dept of Chemistry

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

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

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

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

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.

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)

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

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

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

Piston

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

Example: C3H8(g) + 5 O2(g)→ 3CO2(g) + 4H2O(l)at 298 K  1 atm (1 atm = 101325 Pa), -2220 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

V= nRT/p or Vi = niRT/pex 6 moles of gas: Vi = (6 × 8.314 × 298)/ 101325 = 0.1467 m3 3 moles of gas: Vf = (3 × 8.314 × 298)/ 101325 = 0.0734 m3 work done = -pex × (Vf – Vi) = -101325 (0.0734 – 0.1467) = +7432 J

NB: work done = - pex (nfRT/pex – niRT/pex) = (nf – ni) RT Work done = -∆ngasRT i.e. work done = - (3 – 6) × 8.314 × 298 = + 7432.7 J Can also calculate ∆U ∆U = q +w q = - 2220 kJ w = 7432.7 J = 7.43 kJ  ∆U = - 2220 + 7.43 = - 2212.6 kJ

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

Example: energy diagram

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

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)

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

Dr. S. B Maulage Dept of Chemistry