It's the study of heat energy transfer into or out of the system as it undergoes physical and chemical transformation.
~Limitations~
*only applicable to macroscopic systems as it ignores the internal structure of atoms or molecules. *Concerned only about the initial and final states of the system not the time factor reguarding the rate of reaction.
~Terms and basic concepts~
System: The part of the universe under thermodynamic study Surrounding: Rest of the universe Boundary: Real or imaginary surface separating the system from the surrounding
~Types of thermodynamic systems~
(a) Open system: Both energy and matter can be exchanged with the surrounding
(b) Closed system: Only matter can be exchanged with the system
(c) Isolated system: Neither energy nor matter can be exchanged with the system
Interactions of thermodynamic systems | |||
Type of system | |||
√ | √ | √ | |
× | √ | √ | |
× | × | × |
Heat and work are two ways of transforming energy CH4+2O2→CO2+2H2O
When combustion takes place heat is generated and it can be arranged to obtain mechanical work as well. (Heat factor is represented by "q" Work factor is represented by "w" ) This leads to the idea that in going from reactants to products the property internal energy changes by "∆U". U, which is the sum of kinetic and potential energy due to molecular interactions should, strictly speaking, also include the effects of external fields such as potential energy created by the earth's gravitational field. The latter is neglected in applications in chemistry as we are dealing with relatively small systems.
As the quantity of energy is conserved; ∆U = q + w "U" Is a state function which depends on factors such as temperature, concentration, pressure etc. ∆U= U final - U (initial) ∆ represents difference between initial and final states. The ∆ sign is usually used with "U", not with "d" or "q" as they aren't properties of the initial and final state, they are the means of getting into the initial and final state. Therefore "q" and "w" are path functions as they depend on the route taken. The law is usually applied for closed systems.
~Statements of the 1st law of thermodynamics~
* The energy of an isolated system is a constant. ∆Uisolated=0 (q and w may not be 0 but overall q+w=0)
The universe itself is an isolated system. ∆Uisolated=∆Uuniverse = ∆Usystem + ∆Usurrounding =0
∆Usystem= {-∆Usurrounding}
Energy can only be arranged not created or destroyed. * Requirement of U to be a state function is also a statement of the 1st law. * For macroscopic energy transfer ∆U = q + w * For infinitesimal changes dU = dq + dw
~Sign convention~ q is a small amount of energy supplied to the system and may thus be + or -. w is the work done on the system. Considering q; Heat absorbed =+dq {Endothermic reaction H=(+)}; Heat produced =-dq {Exothermic reaction H=(-)}
Now talking about work; sometimes its convenient to talk about the work done by the system which is represented by w’.{-dw} Work =Force x Distance {Say we are lifting a weight ↑ towards (+). But the weight acts ↓ as {-F(z)}.To lift a certain mass we have to do work. Therefore w’(+) Therefore w’ = - Fz*dz }
w’ =∫dw’ = -∫Fz*dz
Force = Pressure x Area F(z) = -Pext*A {- sign as the pressure is ↓ against the ↑ movement of the piston} dw’ =-F(z)*dz =- {-Pext*A} *dz = Pext*{Adz} Adz=dV dw’ =Pext*dV ∫dw’ =∫Pext*dV {As Pext is a constant;}
∫dw’ =Pext∫dV ∫dw’ =Pext*{Vfinal - V(initial)} ∆w’ =Pext* ∆V Work done by the system (expansion) =dw' + (because ∆V is +) Work done on the system(compression) =dw' - (because ∆V is - )
~Reversible isothermal expansion~
As this process is an isothermal process ∆T=0
When a process goes from the initial to the final state in a single step and cannot be reversed it’s said to be irreversible. The system is in equilibrium on its initial and final stage, not stages in between. A reversible process is considered to proceed from its initial state to its final state through an infinite series of infinitesimally small stages. At initial, final and intermediate stages the system is in equilibrium. Pext=Pint
For expansion; w' = ∫Pext*dV w' = ∫Pint*dV PintV=nRT
w' =∫nRT*dV/V w' =nRT*ln [V(f)/V(i)] P(i)*V(i)=P(f)*V(f) V(f)/V(i)=P(i)/P(f)
w' =nRT* ln [P(i)/P(f)] Isothermal compression of an ideal gas can be derived similarly with the sign changed.
~Irreversible isothermal expansion~ w' = ∫Pext*dV w' = P{V(f)-V(i)} w' = P∆V
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