# Enthalpy and work

1. Jul 16, 2015

### Absentee

Ok, I'll get pretty straight-forward.

So, the usual assumptions:
Energy given to the system: > 0
Energy taken from the system: < 0

Enthalpy is defined as:

ΔH = ΔU + pΔV, so

ΔU = ΔH - pΔV

Let's say that during some chemical reaction there is heat produced (Energy taken from the system):

ΔH = -49 kJ

... And a gas was compressed (Energy given to the system):

W = 5 kJ

To calculate the change in internal energy

ΔU = ΔH - pΔV
ΔU = -49 kJ - 5 kJ = -54 kJ

However, this does not make sense to me.
Shoudn't it be -44 kJ by common sense?

Is it the sign convention is intentionally switched in the first term?

Thanks.

2. Jul 16, 2015

### Staff: Mentor

W is the work done by the system on the surroundings. If the surroundings do work on the system, the work is negative. If the gas is compressed, pΔV is negative. So, pΔV = -5 kJ. So the answer should be -44 kJ.

Chet

3. Jul 17, 2015

### DrDu

Sign convention is always nasty. In chemistry one mostly considers W to be the work done on the system, hence $W=-p\Delta V$ and $\Delta U=Q+W$. Engineers (like Chestermiller) are usually more interested in the work a system can do on its surrounding and use $W=p\Delta V$ together with $\Delta U=Q-W$. From your text, I am inclined to assume that you are using the first convention.

4. Jul 17, 2015

### Absentee

Great, thanks!