Thermodynamics: calculating work question

[jybe]

1. Homework Statement

What are the values of q, w, ΔU, ΔH for the following constant pressure process for a system containing 0.596 moles of CH3OH ?

CH3OH(g, 123.0 ºC, 1.00 atm) ⟶ CH3OH(l, 30.0 ºC, 1.00 atm)Molar heat capacity for CH3OH(g), Cp,m = 44.1 J K−1 mol−1

Molar heat capacity for CH3OH(l), Cp,m = 81.1 J K−1 mol−1

Enthalpy of vaporization, ΔvapH = 35.2 kJ mol−1 at 64.7 ºC and 1.00 atmCalculate:

q (done)

w (work)

ΔH (done)

ΔU (change in internal energy)

Homework Equations

w = -Pext(ΔV) -- following the convention that a +ve answer for work means a compression

ΔU = q + work

The Attempt at a Solution

I calculated q and ΔH to be -24188.775 (they are the same because it's constant pressure).

The problem I'm having is calculating work, because I don't have the change in volume. I think there's some sort of variation of pv = nrt involving the change in moles of gas, but I keep getting it wrong.

Can somebody help me out? Thanks

 

[Chestermiller]

In determining the change in volume, it is valid to neglect the volume of the condensed liquid compared to the original volume of superheated vapor. Using the ideal gas law, what is the original volume of superheated vapor?

 

[jybe]

In determining the change in volume, it is valid to neglect the volume of the condensed liquid compared to the original volume of superheated vapor. Using the ideal gas law, what is the original volume of superheated vapor?

The original volume of gas is:
v= nrt/p

v = (0.596 mol)*(0.082058 atmL/(k*mol))*(123 + 173 K) all over 1 atm of pressure

v = 19.367 L initially of gas

 

[Chestermiller]

So, what's $p\Delta V$ equal to?

 

[jybe]

So, what's $p\Delta V$ equal to?

It is equal to 1 atm * 19.367 L

So to convert 19.367 atm*L to joules, I will multiply by 101.325 J

so work is equal to +1962.36 J ?

Edit: so to get the change in internal energy I just simply add work to q

 

[Chestermiller]

It is equal to 1 atm * 19.367 L

So to convert 19.367 atm*L to joules, I will multiply by 101.325 J

so work is equal to +1962.36 J ?

Edit: so to get the change in internal energy I just simply add work to q

Yes. I would do it using $\Delta H=\Delta U+\Delta (PV)$ (which gives the same answer).

 

[jybe]

Yes. I would do it using $\Delta H=\Delta U+\Delta (PV)$ (which gives the same answer).

Thank you so much! Not sure why I was even having trouble with this.