First Law Quantities Second Law

PHYSICAL CHEMISTRY – I

I C - DR. RANJAN DEY

Calculation of First Law Quantities and Second Law

An infinitesimal change in conditions can reverse the process to restore both system and surroundings to their initial states.

A reversible process is obviously an idealizationLine Integrals

The integral given in earlier expression is

A rev process is one in which system is infinitesimally close to eqb .

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The integral given in earlier expression is called line integral

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Exact and Inexact Differentials

A state function is a property that depends solely on the state of the system. It does not depend on how the system was brought to that state.

When a system is brought from an initial to a final state, the change in a state function is independent of the path followed.

An infinitesimal change of a state function is an exact differential.


Internal energy : state function

: exact differential

, independent of the pathf

f ii

U

dU

dU U U U= − = ∆∫

An infinitesimal change of a state function is an exact differential.

Work and heat are not state functions and do not correspond to exact differentials.

Of the three thermodynamic variables, only two are independent. It is convenient to choose V and T as the independent variables for U.

The First Law

U q w∆ = +

The sum of the heat q transferred to a system and the work w performed on it equal the change ∆∆∆∆U in the system’s internal energy.

According to First law, the total energy for any process,

FIRST LAW OF THERMODYNAMICS

∆E = Closed system


U q w∆ = +Postulate: The internal energy is a state function of the system.

Work and heat are not state functions and do not correspond to exact differentials.

dU dq PdV−= −

∆E =


From First law: ∆U= q + w

( )f

i

V

Vw P V dV=−∫U2 – U1 = q +w

U2 – U1 = qP – P(V2 – V1)qP = U2 +PV2 - (U1 + PV1 )= H2 – H1


P 2 2 1 1 2 1

qP = ∆H constant P, closed system, P -V work only

∆H = ∆ U + P ∆V, constant P

∆ U = qV, closed system, P -V work only, V constant


Heat Capacities

• The heat capacity of a closed system for an infinitesimal process is defined as

• C = dq/dT , dq and dT are heat flowing into the system and temp change of the process.

• For constant pressure process, we get C :


• For constant pressure process, we get C P :• CP = dqP/dT = (∂H/ ∂ T)P

• For constant volume (isochoric) process :• CV = dqV/dT = (∂U/ ∂ T)V

• CP,m = CP/n ; CV,m = CV/n (molar heat cap .) & • cP = CP/m ; c V = CV/m ( specific heat cap .)

In Cyclic process any change in state function is ze ro.∆H = ∆ U = ∆ P = ∆ T = ∆ V = 0In isothermal process, U is constant (Can be achieved by

enclosing system in a thermally conducting walls and place it in constant T bath)

In adiabatic process, dq=0 and q=0(Can be achieved

Calculation of First Law Quantities


In adiabatic process, dq=0 and q=0(Can be achieved by surrounding system with adiabatic walls)

In Constant volume process, work is zero , provided system is capable of doing P-V work only (Can be achieved by enclosing system in rigid walls)

In constant Pressure process, ∆H = qP


Reversible Phase change with constant T&PPhase change or phase transition is a processIn which a new phase appears in a system without chemical Rx.Eg., melting of ice to liquid water, freezing of ice from aq solution.W = -P.dV; ∆H = qP ; ∆ U = q + w.Constant Pressure heating with no phase change

w = w rev = -P.dV ; ∆H = qP ; ∆ U = qP + w Constant Volume heating with no phase change


Constant Volume heating with no phase change

w=0, ∆ U = q + w = q V; ∆H = ∆U + V∆P

Perfect Gas change of state

dU = CVdT , dH = CPdT; w = -∫12 PdV = -∫12 (R/T)dV

Reversible isothermal process in a perfect gas

∆H = 0 and ∆U = 0; w = -nRT ln(V2/V1); q= -w.

Reversible adiabatic process in a perfect gas.

Q=0, w = ∆U; P1V1γ = P2V2

γ RANJAN DEYRANJAN DEYRANJAN DEYRANJAN DEY

Adiabatic expansion of a perfect gas into vacuum

q=0, w=0, ∆H = ∆U + ∆(PV) = ∆U + nR ∆T = 0

The SI unit for q, w, ∆U and ∆H are joules.

• Mechanical Work• Reversible Processes• Reversible Isothermal Expansion/Compression of Idea l Gas• Exact and Inexact Differentials• The First Law• Work and Heat along Reversible Isothermal Expansionfor an Ideal Gas, where U=U(T)

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SECOND LAW

qC is –ve for the heat engine since +ve heat flows out of the system to the cold body.

Efficiency, e, of a heat engine is the fraction of the energy input that appears as useful


of the energy input that appears as useful energy output, that is, appears as work.

e = work output per cycle = -w = |w | energy input per cycle q H qH


e = 1 + qC / qH,

efficiency is less than one as q C is –ve, q H is +ve.CARNOT’S PRINCIPLE

No heat engine can be more efficient than a reversible heat No heat engine can be more efficient than a reversible heat No heat engine can be more efficient than a reversible heat No heat engine can be more efficient than a reversible heat

From I st law: ∆U =0= q +w= qH+qC+w ; -w = q H+qC


No heat engine can be more efficient than a reversible heat No heat engine can be more efficient than a reversible heat No heat engine can be more efficient than a reversible heat No heat engine can be more efficient than a reversible heat

engine when both engines work between the same pair engine when both engines work between the same pair engine when both engines work between the same pair engine when both engines work between the same pair

of reversible heat engines working between the same of reversible heat engines working between the same of reversible heat engines working between the same of reversible heat engines working between the same

pair of temperatures pair of temperatures pair of temperatures pair of temperatures ƬHHHH and and and and ƬCCCC .

The maximum amount of work from a given supply of heat is obtained with

reversible engine


For any closed system undergoing a Carnot cycle, in tegral of dq rev/T around the cycle is zero.

If it is irreversible, it cannot be related to the Carnot cycle and the above condition need not hold.

Since integral of any rev cycle is zero, the value of line integral

ENTROPY


Since integral of any rev cycle is zero, the value of line integral ∫1

2 dq rev/T is independent of path between states 1 and 2 and depends only upon the initial and final states. Hence dq rev/T is the differential of a state function and is defined as entropy, S.

dS ≡ dq rev/T closed system, reversible process.Entropy change in going from state 1 to state 2 is

∆S = S2 – S1 = ∫12 dq rev/T RANJAN DEYRANJAN DEYRANJAN DEYRANJAN DEY

Commonly used units of entropy are J/K or cal/K.Corresponding units of Sm are J/(mol K)

CALCULATION OF ENTROPY CHANGES

For cyclic process : ∆S = 0

Entropy is an extensive state fn


For cyclic process : ∆S = 0For reversible adiabatic process: ∆S = 0 as dq rev = 0 Reversible phase change at constant T and P : ∆S = qrev/T qrev is latent heat of transition at constant T and P. Thus

qrev = qP = ∆H ⇒⇒⇒⇒ ∆S = ∆H/T rev phase change at const T&P

Reversible isothermal process : ∆S = qrev/T Constant Pressure heating : ∆S = ∫T1

T2 CP dT/TRANJAN DEYRANJAN DEYRANJAN DEYRANJAN DEY

1. Rev warming supercooled liquid to 0ºC and 1 atm

2. Reversibly freezing it at 0ºC and 1 atm3. Reversibly cooling the ice to -10ºC and 1 atmMixing of different perfect gases at constant P&T∆S = n Rln(V/V ) + n R ln(V/V ) for irr mixing


∆S = naRln(V/Va) + nbR ln(V/Vb) for irr mixing∆mixS = - naRln(xa) - nbR ln(xb)∆Suniv = ∆Ssys + ∆ssurr

∆suniv =0 for reversible processSuniv = Ssys + suniv remains unchanged in a

reversible process RANJAN DEYRANJAN DEYRANJAN DEYRANJAN DEY

First Law Quantities Second Law

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