Heteroploidy

The Second Law of Thermodynamics

A process that proceeds of its own accord, without any outside assistance, is termed a spontaneous process. The reverse process which doesn’t proceed on its own is referred to as non spontaneous or unnatural process.

In general the tendency to proceed naturally is called the spontaneity.

A spontaneous change is unidirectional and for such processes to occur time is no factor.

Normally it seems like exothermic processes occur spontaneously. But take a cube of ice as example! It would undergo melting when it is left to stand outside, which is an endothermic process. Thus arose the need of inventing another driving force that affects spontaneity. This was known as the entropy change.

What happens in spontaneous processes is that the molecules go into a state of higher randomness than the current state. A change that brings about randomness is more likely to occur than that one brings about order.

Take the example of a ball bouncing ;

When a change occurs the total energy remains constant, according to the first law. The direction of the change is determined by the distribution of energy. In spontaneous processes things tend to go into a state where energy is more chaotically dispersed.

The ordered directional motion of the ball gets converted into the random thermal motion of the floor molecules. As a result the bouncing height decreases as shown above.

On the other way around the ball placed on a hot floor should start bouncing as the reversible process. In order to do so the random thermal energy which is chaotically dispersed should accumulate into the ball performing directional movement.

Reversible processes occur without degrading the quality of the energy

Entropy

Entropy is the thermodynamic state quantity that is a measurement of the randomness or disorder. It’s a state function.

The second law of thermodynamics says that in a spontaneous process the entropy of the universe increases.

∆S_universe ˃ 0

Entropy is related to the heat energy or the degraded energy.

∆S ∞ q

S depends on the temperature as well. If it’s already heated a bit of extra heat won’t help to create much disorder as dramatic as it does when the temperature is very law. Imagine two iron balls that are heated up to 100 ͦ C are placed in two rooms where the room temperatures are 10 ͦ C and 90 ͦ C. Assume that from both balls a q amount of heat is lost to the surrounding. Considering the impact the two energy losses can have on the entropy of the surrounding it appears that the second surrounding’s entropy change will not create any huge impact as the temperature is high and is in state of high entropy. Therefore we can say that;

∆S= q/T or dS= dq/T

∆S is a state function. { S_final-S_initial} and q is a path function. In order to obtain a consistent value ∆S is specified by the reversible pathway. In a reversible path the work done is maximum. As ∆U is constant the q_rev is maximum as well.

∆S= dq_rev /T →(1)

As work done in a reversible path is larger than that of an irreversible path q must be larger as well for the reversible if ∆U is fixed.

dq ^sys _rev ˃ dq^sys_irrev

[dq ^sys _rev]/T ˃ [dq^sys_irrev]/T

∆S^sys ˃ [dq^sys_irrev]/T →Clacius inequality

The entropy change of the universe is that of the system plus that of the surrounding.

dS^uni = dS^sys + dS^surr

dS^uni = dq_rev^sys / T + dq_rev^surr / T

For irreversible processes

Since the surrounding are so large as to be unaffected by the transfer of heat such a transfer can be considered reversible from the point of view of the surrounding.

dq^surr = dq_rev^surr

dS^uni = dq_rev^sys / T + dq^surr / T

In any processe the surroundings gains the heat that is lost from the system. As this is irreversible;

[-dq^sys] = dq^surr

dS^uni = dq_rev^sys / T – dq^sys / T

according to the Clausius inequality;

[dq ^sys _rev]/T ˃ [dq^sys_irrev]/T

Therefore in irreversible/ spontaneous processes dS^uni ˃ 0 (+)

For reversible processes

dS^uni = dq_rev^sys / T + dq_rev^surr / T

[-dq_rev^sys] = dq_rev^surr

dS^uni = dq_rev^sys / T - dq_rev^sys / T= 0

Therefore in reversible processes dS^uni = 0

Thermodynamics

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	Mass flow	Work	Heat
Open	√Y	√Y	√Y
Closed	×N	√Y	√Y
Isolated	×N	×N	×N

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. ∆U_isolated=0 (q and w may not be 0 but overall q+w=0)

The universe itself is an isolated system. ∆U_isolated=∆U_universe = ∆U_system + ∆U_surrounding =0

∆U_system= {-∆U_surrounding}

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’ = - F_z*dz }

w’ =∫dw’ = -∫F_z*dz

Consider the following piston; In a certain moment the gas expands making the piston move upwards against the external pressure.

Force = Pressure x Area F_(z) = -P_ext*A {- sign as the pressure is ↓ against the ↑ movement of the piston} dw’ =-F_(z)*dz =- {-P_ext*A} *dz = P_ext*{Adz} Adz=dV dw’ =P_ext*dV ∫dw’ =∫P_ext*dV {As P_ext is a constant;}

∫dw’ =P_ext∫dV ∫dw’ =P_ext*{V_final - V_(initial)} ∆w’ =P_ext* ∆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. P_ext=P_int

For expansion; w' = ∫P_ext*dV w' = ∫P_int*dV P_intV=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' = ∫P_ext*dV w' = P{V_(f)-V_(i)} w' = P∆V