Question

Which one has a larger final entropy, reversible isothermal process or reversible adiabatic process when expanding to the same final volume for an ideal gas?

What do you assume the entropy S for a perfect crystal when T = 0?

Is the S(0) still zero if the crystal is not perfect? Why?

How to calculate dS when under the constant pressure and c_p is given?

How to calculate the entropy of H₂O at 500 K?

How to calculate the entropy change for a chemical reaction if you can find the S for each chemical?

Answer 1

question -1 :

first we have to understand meaning of entropy for the process.entropy is define as "the measure of a system’s thermal energy per unit temperature that is unavailable for doing useful work. in other word we can says that entropy is measure of the molecular disorder, or randomness, of a system "

two process are given.one is reversible isothermal process and other one is reversible adiabatic process  for ideal gas which have same final volume.

reversible isothermal process :

for reversible isothermal process, isothermal process means temperature of system remains constant.it means ΔT = 0 for the isothermal process.

Isothermal processes are especially convenient for calculating changes in entropy since, in this case, the formula for the entropy change, ΔS

where Q_rev is the heat transferred reversibly to the system and T is absolute temperature.This formula is valid only for reversible isothermal process; that is a process in which equilibrium is maintained at all times

The entropy is a state variable so the entropy change of the system is the same as before. In this case, however, heat is transferred to the system from the surroundings ( ) so that

The heat transferred from the surroundings, however, is equal to the heat received by the system: .

The total change in entropy (system plus surroundings) is therefore

The reversible isothermal process has zero total change in entropy.

reversible adiabatic process :

for reversible adiabatic process there is no mass and energy transfer occured.so net change in internal energy is equal to zero. Q_rev = 0

reversible adiabatic processes are especially convenient for calculating changes in entropy since, in this case, the formula for the entropy change, ΔS

here ,  Q_rev = 0 , so entroy for the reversible adiabatic process is equal to zero

so answer is both process have zero entropy.reversible isothermal process or reversible adiabatic process  have zero entropy

question-2 :

Third Law of Thermodynamics says: "The entropy of a pure perfect crystal is zero at T = 0 (0° K).the Third Law of Thermodynamics means that as the temperature of a system approaches absolute zero, its entropy approaches a constant (for pure perfect crystals, this constant is zero)

question:3

A pure perfect crystal is one in which every molecule is identical, and the molecular alignment is perfectly even throughout the substance. For non-pure crystals, or those with less-than perfect alignment, there will be some energy associated with the imperfections, so the entropy cannot become zero.

for non pure crystal , entropy cannot become zero.

question-4:

for constant pressure , a process is said to be isobaric process. if C_p is given ,

then dS = ΔQ / T

but  ΔQ for constant pressure is given by,

ΔQ = n C_p dT

dS = ΔQ / T

= n C_p dT / T

dS = n C_p ln ( T₂ / T₁ ) { ln T₂ - ln T₁ = ln ( T₂ / T₁ ) }

here, T₂ = final temperature n = gram / mole

T₁ = initial temperature

C_P = Specefic heat at constant pressure

question-5:

T = 500 k

entropy of H₂O = Q / T

latent heat of condensation of water is 40.7 kJ/mole of water. so Q = 40.7 kJ/mole = 40.7 x 10³ J/mole

T = 500 K

S = 40.7 X 10³ / 500

S = 81.4 J / mole K

question-6 :

We return to considerations of a chemical reaction at p,T = const, which we represent as

here process occured between two reactor A and B and we get product of C and D

or

or

The reaction entropy is the entropy change per mole of reaction as written.

If we wish to break up the entropy change into contributions from individual species, we have to consider the special case of reactants in their standard states being converted to products in their standard states. Then we can write

or

is the standard molar entropy of compound J.