Unlike the internal energy and heat content ,the work obtainable when a system undergoes change depends not only on the initial and final states but also on how the change takes place .
The work performed by a system can vary all the way from zero, “in case of expansion into vacuum ,up to a maximum under specified conditions .
•Any process in which the driving force is only infinitesimally greater than the opposing force ,and which canbe reversed by increasing the opposing force by an infinitesimal amount constitute reversible process .
•All naturally occurring process are irreversible
•Reversibility can be approached in galvanic cells .
•When the work is performed isothermally and reversibly, then the work Wm is done at the expense of the maximum work content of the system.
By applying pressure on the piston on the left slowly enough so as not to change the pressure p1, a volume of gas v1 was forced slowly through the pours plug and allowed then to expand to the pressure p2 and volume v2by moving the piston on the right. 
Under these conditions joule - Thomson observed near room temperatures that all gases show cooling except hydrogen actually became warmer, the extent of the temperature change was found to depend on the initial temperature and pressure of the gas.
•The magnitude of the observed effects controlled by the joule – Thomson coefficient µ which is defined as:
‘The number of degrees temperature change produced by atmosphere drop in pressure under constant enthalpy”.
Hydrogen and helium can be cooled adiabatically if they are first brought to a sufficient low temperature, this temperature is called inversion temperature, namely, the temperature at which the gas exhibit neither cooling nor heating.
This effect is of great importance in liquefaction of gases.
The carnot cycle
The work performed by a system can vary all the way from zero, “in case of expansion into vacuum ,up to a maximum under specified conditions .
•Any process in which the driving force is only infinitesimally greater than the opposing force ,and which canbe reversed by increasing the opposing force by an infinitesimal amount constitute reversible process .
•All naturally occurring process are irreversible
•Reversibility can be approached in galvanic cells .
Reversibility and maximum work
•The amount of work a system has to perform to give certain change depends on the opposition the system experiences to the change, the greater that resistance is, the more work must be done by the system.
•Consider the expansion of an ideal gas against a pressure P through an infinitesimal volume change dv, the work done is zero , if P is zero, however as P increased more and more work has to be done as the pressure approaches that of the gas ifself.
•It is evident that the work which may performed by a system is a maximum when the opposing pressure P differs only infinitesimally from the internal pressure of the gas itself. Thus maximum work is obtainable only from a system when the change taking place in it is entirely reversible.
since w = ∫ Pdv
for reversible process the external pressure is at all times only infinitesimally lower than the pressure of the gas itself, we may substitute P = nRT/ v
So Wm= ∫v1v2 nRTdv/v
• Wm = nRT ∫v1v2 dv/v = nRT ln v2/v1
alternatively Wm = nRT ln p1/p2
Maximum work of adiabatic expansion.
The expression in question is arrived at as:
Differentiate the expression pvγ = constant.
γ pvγ-1 dv + vγdp = 0 dividing by vγ-1
γ pdv + v dp = 0
v dp = - γ pdv
complete differentiation of pv = nRT gives:
P dv + v dp = nRT
On substitution of the expression for v dp
P dv - γ P dv = nR dT
P dv ( 1- γ) = nR dT
P dv = nRdT / ( 1- γ)
Wm = ∫v1v2 p dv = ∫T1T2 nR dT / ( 1- γ)
= nR ( T2 – T1) / ( 1- γ)
T2> T1 Wm is negative and work is done on the gas, when work is done by the system
T2< T1 and Wm is positive.
The maximum work function A
•We have seen that the amount of work obtainable from a process depends on the manner in which the work has been performed. Under isothermal and reversible conditions a definite path of passing from the initial to the final state along which the maximum work done is definite and dependent only on the two states.
•At constant temperature, when the maximum work being a function of the states as the internal energy E and heat content H.
•Therefore , we may think of a system as possessing in each state a certain amount of maximum work content A, called (the Helmholtez free energy), when the system passes from one state to another, the change in the work function ∆ A is equal;
∆ A = A2 - A1
∆ A > 0 means increase in the work content
∆ A < 0 means decrease in the work content
•When the work is performed isothermally and reversibly, then the work Wm is done at the expense of the maximum work content of the system.
Thus Wm = - ∆A
From the first law; ∆E = qr - Wm
= qr + ∆A
The joule – Thomson effect
•For an ideal gas, pv product is a constant , if this gas expands under adiabatic conditions into a vacuum, then;
q = 0 w = 0 ∆E = 0
i.e ∆E for the gas remains constant as vp and consequently the temperature. This is equivalent to say” At constant temperature the internal energy of an ideal gas is independent of the volume the gas occupies or ( ∂ E / ∂v)T = 0
With real gases the situation is different;
Joule – Thomson experiment
At left piston At right piston
Work done is – p1v1 work done is p2v2
The net work w = p2v2– p1v1
q = 0 since the process is adiabatic
∆E = E2 – E1 = - w = - (p2v2– p1v1 )
E2+ p2v2 = E1 + p1v1
H2 = H1 ∆H = 0
i.e. the process was conducted at constant enthalpy. Under these conditions joule - Thomson observed near room temperatures that all gases show cooling except hydrogen actually became warmer, the extent of the temperature change was found to depend on the initial temperature and pressure of the gas.
•The magnitude of the observed effects controlled by the joule – Thomson coefficient µ which is defined as:
‘The number of degrees temperature change produced by atmosphere drop in pressure under constant enthalpy”.
µ = (∂T / ∂p)H
µ positive cooling
µ negative heating
Hydrogen and helium can be cooled adiabatically if they are first brought to a sufficient low temperature, this temperature is called inversion temperature, namely, the temperature at which the gas exhibit neither cooling nor heating.
This effect is of great importance in liquefaction of gases.
•A question of great significance to how extent heat is convertible to other forms of energy which may utilize to do work, experience has shown that operating heat engines which absorb heat at temperature T2 and reject the waste heat at a lower temperature T1, can convert only a fraction of the absorbed heat into work.
Division of equation (A) by q2 yields:
Equation (B) can be simplified,since the points (p1v1) and (p3v3) lie on the same adiabatic ,
Similarly, the points (p2v2) and (p4v4) lie on the same adiabatic then:,
On dividing the first equation by the second we obtain: v1 /v2 = v3 /v4 consequently
substituting this value in equation (B) we get
•Thus the maximum work that may be recovered from a system during a reversible cyclical process is equal to the heat absorbed multiplied by the ratio (T2-T1/T2 )
•The ratio of wm /q2 is called the thermodynamic efficiency of the process.
•Thermodynamic efficiency must be the same for all the process operating under the given temperature conditions.
•Theoretical consideration show that even an ideal engine , operating under ideal conditions, would be able to convert only a certain fraction of the absorbed heat into work and that this fraction depend only on the operating tempe.
•To establish the above deductions, let us consider the sequence of operation called a carnot cycle, which consists of four steps, two isothermal and two adiabatic.
W1 = nRT2 ln v1/v2 = q2 (a)
W2 = - ∆E = - ncv ( T1 –T2) (b)
W3 = nRT1 ln v4/v3 = - q1 ( c)
W4 = - ∆E = -ncv (T2 – T1)= ncv (T1- T2) (d)
Wm = w1 + w2 + w3 + w4
= nRT2ln v2 /v1 + nRT1 ln v4/v3 = q2-q1 (A)
Wm/q2 = q2 – q1 /q2 =
nRT2 ln v1/v2 + nRT1 ln v4/v3 / nRT2 ln v1/v2(B)
Equation (B) can be simplified,since the points (p1v1) and (p3v3) lie on the same adiabatic ,
then: v1T2cv/R = v3T1cv/R
Similarly, the points (p2v2) and (p4v4) lie on the same adiabatic then:,
V2 T2cv/R = v4T1cv/R
On dividing the first equation by the second we obtain: v1 /v2 = v3 /v4 consequently
ln v4 /v3 = - ln v1 /v2
substituting this value in equation (B) we get
W/q2 =RT2 ln v1 /v2-RT1 ln v1 /v2/RT2ln v1 /v2
= (T2 –T1) /T2
and wm = q2 ( T2-T1 / T2)
•The ratio of wm /q2 is called the thermodynamic efficiency of the process.
•Thermodynamic efficiency must be the same for all the process operating under the given temperature conditions.
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