# What will be the sign of work done?

Consider the following question:

The enthalpy of vapour is $$10\ \mathrm{kcal}/\mathrm{mol}$$ higher than liquid at $$350\ \mathrm{K}$$ and $$2\ \mathrm{atm}$$. Under given conditions, the difference in internal energy of vapour and liquid will be:

(a) $$10700\ \mathrm{cal}$$

(b) $$-9300\ \mathrm{cal}$$

(c) $$9300\ \mathrm{cal}$$

(d) $$-10700\ \mathrm{cal}$$

My solution is as follows:

Let the internal energy of the liquid be $$U_\mathrm L$$ and that of vapor be $$U_\mathrm G$$. We are required to calculate $$U_\mathrm G - U_\mathrm L$$. As per the question,

$$\because H_\mathrm G = U_\mathrm G + W$$, where $$H_\mathrm G$$ is enthalpy of vapor.

$$10000 + U_\mathrm L = U_\mathrm G + W$$

Where $$W$$ is the work done due to the expansion of vapor as the liquid changes state. The magnitude of work done $$|W| = \Delta n_\mathrm gRT = 1 \times 2 \times 350 = 700\ \mathrm{cal}$$.

Now the real question starts, what is the sign of $$W$$? As per the IUPAC convention, work of expansion is negative, hence $$W = -700\ \mathrm{cal}$$. So our answer will come out to be,

$$10000 + U_\mathrm L = U_\mathrm G - 700$$ $$\implies 10000 + 700 = U_\mathrm G - U_\mathrm L$$

$$\therefore U_\mathrm G - U_\mathrm L = 10700\ \mathrm{cal}$$

But the terse solution given to this question is:

As per the solution $$W = +700\ \mathrm{cal}$$, i.e. they have considered the work of expansion to be positive.

Where are we wrong? Or are we right?

Analyzing the problem by considering the work done is not the easiest place to start solving this problem. It's easier to consider the definition of enthalpy in terms of internal energy and $$pV$$: $$\Delta H \equiv \Delta U + \Delta(pV)$$ It follows that $$\Delta U = \Delta H - \Delta(pV)$$ Taking $$p$$ as constant and equal to $$\pu{2 atm}$$ $$\Delta U = \Delta H - p(V_g-V_l)$$ Assuming $$V_l$$ to be much smaller than $$V_g$$, $$\Delta U = \Delta H - pV_g$$ You can equate $$-pV_g$$ with the work $$W$$ required to expand the gas. This is consistent with the usual convention in chemistry: work of expansion is negative (as you write). The correct result follows.
On a side note, since work represents a change in energy, not a state function, this equation should be written in terms of changes: $$\Delta H = \Delta U - w$$. Also note that this equation is only valid for the condition of constant pressure.