Calculate the change in entropy for the system for each of the following cases.

Question

Calculate the change in entropy for the system for each of the following cases. Explain the sign that you obtain by a physical argument. A gas undergoes a reversible, adiabatic expansion from an initial state at 500 K, 1 MPa, and 8.314 L to a final volume of 16.628 L. One mole of methane vapor is condensed at its boiling point, 111 K; Delta H^vap = 8.2 kJ/mol One mole of liquid water is cooled from 100 degree C to 0 degree C. Take the average heat capacity of water to be 4.2 J/gK. Two blocks of the same metal with equal mass are at different temperatures, 200 degree C and 100 degree C. These blocks are brought together and allowed to come to the same temperature. Assume that these blocks are isolated from their surroundings. The average heat capacity of the metal is 24 J/mol K.

Explanation / Answer

a) Reversible adiabatic process refers to isentrpoic process and entropy change= 0

b) Entropy change for one mole of gas= delHVap/T = 8.2/111=0.0738 Kj/K

Vaporization is spontaneous process

c) Entropy change= moles* molecular weight of water* Specifc heat of water*ln(T2/T1)

T2= 0 deg.c= 0+273.15= 273.15K and T1= 100deg.c= 100+273.15= 373.15K

Entropy change= 1*18*4.18**ln(273.15/373.15)= =23.4718 J/K

this is not a spontaneous process

d) Heat lost by metal block= heat gained by metal block ( since the blocks are isolated from surroundings)

m*cp* ( 200-t)= m*Cp*(100-t)

m= moles of of metal and Cp= specific heat, t= equilibrium temperature deg.c

T= (200+100)/2= 150

Entropy change of hot block= m*24*ln({150+273.15)/(200+273.15)}=-2.68m

Entropy change of cold block= m*24*ln {(150+273.15)/ (100+273.15)}=3.02m

total entropy change= entropy change of hot block + entropy change of cold block= (3.02-2.68)m=0.34m

this is spontaneous process.

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