please answer all parts and label each part clearly. thanks One very useful prop
ID: 540107 • Letter: P
Question
please answer all parts and label each part clearly. thanks
One very useful property of Maxwell Relationships is they apply to all materials, not just gases. In this problem, we'll use some of these relationships to calculate the change in temperature that occurs when liquid water is reversibly expanded. a. Write down an expression for the full differential of the entropy, taking it to be a function of pressure and temperature (i.e. write down dS for S(T,P)). Using a Maxwell Relationship and our thermodynamic definition of the entropy, show that this expression simplifies to: ov, if we consider a reversible adiabatic process. (Yes, this is tricky, but I know you can do it!) 1 (av, b. The quantity describes the fractional change in the volume of a system when it is heated by a given amount. This quantity is known as the isobaric (constant pressure) thermal expansion coefficient and often given the symbol . For many compounds, the value of is tabulated and this quantity can be assumed to be constant over a set temperature range. For liquid water at 25 °C, = 2.1 x 10-4 K1 Calculate how much the temperature of liquid water changes when it is expanded adiabatically and reversibly from an initial pressure of 101 atm to a final pressure of 1 atm. The temperature of the liquid is initially at 25 oC. You may assume that water's molar volume is roughly constant throughout this expansion.Explanation / Answer
S= S(T,P)
dS= (dS/dT)PdT+(dS/dP)TdP, since deltaH= T*dS+V*dP, (dH/dT)P= T*(dS/dT)P, (dS/dT)P= CPm/T, Cpm= mean specific heat
from maxwell relation, (dS/dP)T =-(dV/dT)P
hence dS= CPmdT/T - (dV/dT)PdP
for adiabatic reversible process, dS= 0, Cpm*dT/T- (dV/dT)PdP=0 (1)
but given alpha= 2.1*10-4/K, (dV/dT)P= 2.1*10-4*Vm=2.1*10-4*0.0188 L/mole =3.9*10-6 L/mole/atm
hence Eq.1 becomes 4.184 J/gm.deg.K* dT/T- 3.9*10-6dP =0
when integrated
4.184* ln (T2/T1)= 3.9*10-6*(101-1) L.atm*101.3 J=0.0399
ln(T2/298)= 0.0399
T2/298= 1.041
T2= 1.041*298 =310K
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