1. Calculate the change in entropy when 27 kJ of energy is transferred reversibl
ID: 900621 • Letter: 1
Question
1. Calculate the change in entropy when 27 kJ of energy is transferred reversibly and isothermally as heat to a large block of iron at (a) 10°C, (b) 50°C.
2. Calculate S (for the system) when the state of 2.00 mol of perfect gas atoms, for which Cp,m = 5/2R,is changed from 20°C and 1.00 atm to 125°C and 4.00 atm. How do you rationalize the sign of S?
3. Consider a system consisting of 2.0 mol CO2(g), initially at 25°C and 12 atm and confined to a cylinder of cross-section 10.0 cm2. It is allowed to expand adiabatically against an external pressure of 1.0 atm until the piston has moved outwards through 20 cm. Assume that carbon dioxide may be considered a perfect gas with CV,m = 28.8 J K1 mol1 and calculate (a) q, (b) w, (c) U, (d) T, (e) S.
Explanation / Answer
1- delta S = Q /T
a)= 27 x 10^3 J/ 283K = 95.406J/K
b) delta S=27x10^3 J/323K = 83.59 J/K
2-n= 2mol , Cpm= 5/2R
T1= 293K , T2 = 398K
P1=1atm , P2 = 4atm
TdS =dH - VdP
dS = dH/T -V/T dP
from ideal gas law
PV=RT=> V/T = R/P
dH/T =Cp => dH=CpdT
putting these value in above equation & on interigation
dS = Cp/T dT - R/P dP
delta S = Cp ln T2/T1 - R ln P2/P1 = 5/2R ln 398/293 - R ln 4/1 = 5/2R x 0.3060 - R x 1.38=1.53R - 2.76R=-1.23R
S = n*s = 2x -0.0821x1.23 =- 0.201966
intropy is meaer of disoderness.-ve sign indicate decrease in intropy.
3-1st law energy equation ,
U = 1W2.=PdV =
Related Questions
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.