Liquid methanol is heated from 25^{o}C to 100^{o}C in the pistonandcylinder device shown below. The initial pressure is 100 kPa and the spring causes the pressure to increase during the process to 200 kPa. 

















Calculate DU and DH in J/mol. Assume C_{P} is a constant and has a value of 83.4 J/molK. Assume the molar volume is also constant and has a value of 0.01848 mol/L. 

















Read : 
Construct a good HPP for this process. Treat the liquid methanol as an incompressible fluid and verify this assumption at the end. This will simplify determining ΔU and ΔH for changes in pressure. Use the given heat capacity to determine ΔU and ΔH for changes in temperature. 




















Diagram: 





















Given: 
T_{1} 
25 
^{o}C 



T_{2} 
125 
^{o}C 






P_{1} 
100 
kPa 



P_{2} 
200 
kPa 






C_{P} 
83.4 
J/molK 



V 
0.01848 
mol/L 




















Find: 
ΔU_{12} 
??? 
J/mol 



ΔH_{12} 
??? 
J/mol 




















Assumptions: 
1  Liquid methanol is incompressible. The molar volume is constant throughout this process. 




















Equations
/ Data / Solve: 




























The reason we use a hypothetical process path is to
break a complex process into a series of simpler steps. 






In this problem, step 1A is isobaric and step A2 is isothermal. 






Because U and H are state variables, they are additive, as follows. 






















Eqn 1 



Eqn 2 





















ΔH for Step 1A can be determined as follows because the heat capacity is a constant. 



























Eqn 3 





















Plugging values into Eqn 2 yields: 


ΔH_{1A} 
8340.0 
J/mol 





















Now we can use the
definition of enthalpy to
help us determine ΔU. 
























Eqn 4 





















Since P_{A} = P_{1}, ΔP = 0. 






We boldly assumed the molar volume of the liquid methanol was constant throught this process, ΔV_{1A} = 0. 





















The result is easy to
compute! 


ΔU_{1A} 
8340.0 
J/mol 





















For an incompressible liquid, as we have
assumed liquid methanol to be, U is not a function of P. 





















Therefore : 





ΔU_{A2} 
0 
J/mol 





















To determine ΔH, we must return to the
definition of enthalpy. 
























Eqn 5 





















Plugging values into Eqn 5 yields : 


ΔH_{A2} 
1.848 
J/mol 





















You can see that ΔH_{A2} is very small compared to ΔH_{1a}. It is often
neglected unless the change in T is very small or the change in P is very large indeed. 





















Now, we can plug
values back into Eqn 1 and Eqn 2 to complete the solution of
this problem. 





















ΔU_{12} 
8340 
J/mol 



ΔH_{12} 
8342 
J/mol 




















Verify: 
We cannot verify the incompressibility of liquid methanol using only the information given in the problem statement. 





















However, the NIST Webbook yields the following
data for the molar volume of liquid methanol. 





















T (C) 
P (MPa) 
V (L/mol) 

T (C) 
P (MPa) 
V (L/mol) 








25 
0.1 
0.017358 

25 
0.2 
0.018068 








100 
0.1 
0.018479 

100 
0.2 
0.018797 























The data show that the
molar volume changes by about 6% during Step 1A and about 4% during Step A2. 






This seems like a lot
of error, but it does not translate into as much error in ΔU or ΔH. 





















The first place this
assumption matters is in determining ΔU_{1A}. 


























Eqn 6 






















Eqn 7 


P ΔV_{1A} 
0.071 
J/mol 





















This is less than 1% of ΔH_{1A} ! This is not
significant. 





















The next place the
incompressibility assumption matters is in ΔH_{A2}. 
























Eqn 5 



























P ΔV_{A2} 
0.0636 
J/mol 





















This is less than 4% of ΔH_{A2} but it is less than
1% of ΔH_{12} ! This is not significant either. 





















So for determining
changes in U and H the constant molar volume assumption for liquid methanol
was reasonable for these conditions. 





















When solving a problem
for chemical that is in the NIST Webbook database, you should
use the best available information. 





















But in the absence of
extensive data about the molar volume of liquids, it is very common and often accurate to assume they are incompressible over pressure ranges of 1 MPa
or even more. 




















Answers : 
ΔU_{12} 
1.85 
J/mol 



ΔH_{12} 
8340 
J/mol 





















































