Example Problem with Complete Solution

3D-2 : Heating Liquid Methanol in a Piston-and-Cylinder Device 3 pts
Liquid methanol is heated from 25oC to 100oC in the piston-and-cylinder 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 CP is a constant and has a value of 83.4 J/mol-K. 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.
Given: T1 25 oC T2 125 oC
P1 100 kPa P2 200 kPa
CP 83.4 J/mol-K V 0.01848 mol/L
Find: ΔU12 ??? J/mol ΔH12 ??? 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 1-A is isobaric and step A-2 is isothermal.
Because U and H are state variables, they are additive, as follows.
Eqn 1
Eqn 2
ΔH for Step 1-A can be determined as follows because the heat capacity is a constant.
Eqn 3
Plugging values into Eqn 2 yields: ΔH1A 8340.0 J/mol
Now we can use the definition of enthalpy to help us determine ΔU.
Eqn 4
Since PA = P1, ΔP = 0.
We boldly assumed the molar volume of the liquid methanol was constant throught this process, ΔV1A = 0.
The result is easy to compute! ΔU1A 8340.0 J/mol
For an incompressible liquid, as we have assumed liquid methanol to be, U is not a function of P.
Therefore : ΔUA2 0 J/mol
To determine ΔH, we must return to the definition of enthalpy.
Eqn 5
Plugging values into Eqn 5 yields : ΔHA2 1.848 J/mol
You can see that ΔHA2 is very small compared to ΔH1a. 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.
ΔU12 8340 J/mol ΔH12 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 1-A and about 4% during Step A-2.
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 ΔU1A.
Eqn 6
Eqn 7 P ΔV1A 0.071 J/mol
This is less than 1% of ΔH1A ! This is not significant.
The next place the incompressibility assumption matters is in ΔHA2.
Eqn 5
P ΔVA2 0.0636 J/mol
This is less than 4% of ΔHA2 but it is less than 1% of ΔH12 ! 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 : ΔU12 1.85 J/mol ΔH12 8340 J/mol