1.9 ENTROPY CHANGE IN A IRREVERSIBLE PROCESS – B.SC 2ND YEAR.

Let us now suppose that the same system undergoes the same change in an irreversible manner at a constant temperature T. Entropy being a state function , the increase in entropy of the system will be the same irrespective of the fact whether the change has occurred reversibly or irreversibly.

Thus, ΔS will remain the same as before and represented by the expression,

ΔSsystem = qrev/T …(i)

However, actual amount of heat absorbed by the system from the surroundings will be less, being equal to qirr

It also follows from here that heat lost by the surroundings will be less than qrev.However, in view of the infinitely large size of the surroundings, the lost heat by the surroundings is supposed to take place reversibly , i.e. , infinitelyslowly . therefore, mathematically the entropy change of the surroundings is given by the expression,

ΔSsurroundings = – qirr/T …(ii)

The total entropy change for the combined system and the surroundings ,is,therefore,given by ,

ΔSsystem + ΔS surroundings = qrev/T – qirr/T

Now, since work done in reversible process is maximum , therefore, we can say that

wrev > wirr(iv)

Also, as internal energy (U) is a state function ,therfore, the value of ΔU remains the same whether the process is caried out in a reversible or an irreversible manner.

As such we can say that,

ΔU= qrevwrev = qirrwirr(v)

From equation (iv) and (v) we can say that

qrev > qirr

or,

qrev/ T > qirr/T

or,

qrev/T – qirr/T > 0 …(vi)

where ‘T‘ is the temperature at which the change takes place.

From equation (iii) and (vi), we can say that,

ΔSsystem + ΔSsurroundings > 0

i.e , Entropy change for the combined system and the surroundings in an irreversible process is greater than zero , In other words , an irreversible process is accompained by a net increase of entropy.

It may, therefore be concluded from above discussion that entropy of the system and its surroundings taken together increase in a thermodynamically irreversiblr process at constant temperature but remains unchanged in a thermodynamically reversible process

Thus, for a thermodynamically reversible process

ΔSsystem + ΔSsurroundings = 0 and,

for an thermodynamically irreversible process

ΔSsystem + ΔSsurroundings > 0

on combining both the results , we can say that

ΔSsystem + ΔSsurroundings ≥  0

where sign ‘equal to‘ refers to a reversible process while the sign ‘greater than‘ refers to an irreversible process

The above generalizations can also be used to distinguish between a thermodynamic reversible and an irreversible process.

Second law of Thermodynamics in term of entropy change – Entropy of the Universe : since all process in nature are occurring spontaneously and are thermodynamically irreversible, it follows, therefore. that entropy of the universe is continously increasuing . This is yet another statement of Second law of Thermodynamics.

Clausius summed up the essentials of first law and second law of thermodynamics as :

The energy of the universe remains constant but entropy of the universe is continously increasing and tends towards a maximum.

1.5 THERMODYNAMIC SCALE OF TEMPERATURE

BEFORE PRECEEDING…, MOTIVATION IS NECESSARY

IF IT DON’T CHALLENGE YOU, IT DON’T CHANGE YOU
  • As we know the scale of an thermometer is depends on the physical properties like “Thermal expansion” of the liquid used in the thermometer.
  • Since,  different liquids have different cofficient value of expansion, and also do not expand uniformly over a given range of temperature, therefore thermometers are constructed by using different liquids will show different numerical values on thermometer
  • THE TEMPERATURE AT WHICH AN IDEAL GAS CEASES TO EXSIST IS TAKEN AS ZERO DEGREE ABSOLUTE (-273.15ᵒC)  AND THE SCALE DEVELOPED BY TAKING ZERO DEGREE ABSOLUTE AS THE ZERO  IS KNOWN AS THE IDEAL GAS TEMPERATURE SCALE AND IS INDEPENDENT OF THE IDEAL GAS TAKEN.

LORD KELVIN, developed a scale of temperature based upon the SECOND LAW OF THERMODYNAMICS

  • Based on his temperature scale and efficiency of the reversible heat engine, He named his temperature scale THERMODYNAMIC SCALE OF TEMPERATURE Or KELVIN SCALE OF TEMPERATURE. it is very similar to the ideal gas temperature scale.
  • All the reversible engines are operating between tempratures T₁ and T₂ having the similar efficiency i.e. , η = f ( T₁ – T₂ )
  • where f is a universal function independent of the working fluid.
  • since,
    • η = q₂q₁/q₂ = 1 – q₁/q₂ = 1 + q₂/q₁,
  • therefore, q₂/q₁ must also be a function of temperatures of the two reservoirs. A number of different temperature scale that satisfy this relation can be deviesed ; but by taking temperature as directly proportional to the ideal gas temperature scale.
  • If q₂ is the heat transfer from the source (heat resevior at higher temperature) and q₁ is the heat transfer to the sink (reservoir at lower temperature) and θ₁ and θ₂ are their temperatures on kelvin scale, then
    • q₂/q₁ = θ₁/θ₂
  • in other words , we have defineed the ratio of two temperatures in a way is independent of nature of the thermometric substance.
  • Taking the reciprocal of equation 1 , we have
    • q₁/q₂ = θ₁/θ₂ or 1- q₁/q₂ = 1- θ₁/θ₂
    • q₂q₁/q₂ = θ₂-θ₁/θ₂
  • Now, if θ1 = 0 (which corresponds to zero of nwe scale), then
    • q₂-q₁/q₂= θ₂/θ₂= 1
  • In other words, this means that zero on the kelvin scale is the temperature of the sink for a reversible engine whose efficiency is unity but according to equatiion (15) section 1.3,
    • Efficiency(η) = q₂-q₁/q₂ = T₂ – T₁/T₂
  • it is possible only at absolute zero on the ideal gas scale of temperature. from this, it can be concluded that kelvin scale and gas scale are identical; provided the gas is supposed to be ideal, The temperature on the thermodynamic scale is , therefore expressed in kelvin (K).

Learning fact

The only two non-silvery metals are gold and copper. …

ASSESSMENT – 1.

FOLLOW – UP PROBLEMS

1. Calculate the amount of heat supplied to carnot cycle working between 368K and 268K, if the maximum work obtained is 890 joules.                                   

2. A carnot engin eworking between 373K and 273K takes up 840 joules from the high temperature reservoir. Calculate the work done , the heat rejected and the efficiency of the heat engine.

3. Calculate th efficiency of a steam engine which operates between 373K and 573K. What is the minimum heat which must be withdrawn from the reservoir to obtain 171.52 joule of work ?

4. Two steam engines , one red and one black , take steams at 277ᵒC, while red engine rejects it at 27ᵒC , the  black one reject it at 47ᵒC. compare the efficiencies of the engines.   

5. Calculate the efficiencies of a carnot engine working between 200K and 1000K.                              

6. Heat supplied to a carnot engine is 453.6 kcal. How much useful work can be done by the engine which work between 0ᵒC and 100ᵒC ? 

7. A carnot engine converts one sixth of heat input into work . When the temperature of the sink is lowered to 335K, its efficiency is doubled. Calculate the temperature of source and the sink.                       

8. If the efficiency of a carnot engine is the same (i) between 100 K and 400 K and (ii) between T K and 800 K, calculate the temperature T of the sink.      

9. Calculate the maximum efficiency of a steam engine operating between 110ᵒC and 250ᵒC. What would be the  efficiency of engine if the boiler temperature is raised to 140ᵒC, the temperature of the sink remaining same?

10. What per cent T₁ and T₂ for a heat engine whose efficiency is 10% ?