Q:

Unit confusion - How Pressure and Volume combines to Joule

Imagine some helium in a cylinder with an initial volume of $1$ liter
  ant initial pressure of $1$ atm. Somehow the helium is made to expand
  to a final volume of $3$ liters, in such a way that its pressure rises
  in direct proportion to its volume. Calculate the work done on the gas during this process, assuming there are no other types of work being done.

I know the following:

$\text{J}=\text{Pa}\cdot\text{m}^3$
$1 \ \text{atm} = 1.013\cdot10^5 \ \text{Pa}$
$1 \ \text{litre} = (0.1 \ \text{m})^3$

I want to compute work done during quasistatic compression. I do get the correct integral, where the formula is
$$W=-\int\limits_{V_\text{initial}}^{V_{\text{final}}}P(V) \ dV,$$
where $P(V)=V$ in my case. But the answer should be $-400 \ \text{J}$ and not $-0.4 \ \text{J}$ like I'm getting. I assume I'm doing something wrong with the units here:
$$W=-\int\limits_{1 \ \text{litre}}^{3 \ \text{litres}} V \ \text{atm} \ dV=-\int_\limits{0.001\text{m}^3}^{0.003\text{m}^3}V\cdot1.013\cdot10^5\ \text{Pa} \ dV=-0.4052 \ \text{J}.$$
Can someone show me how to get the correct answer?

A:

When you specify units of L for volume and atm for pressure, then when you say $P(V)=V$, what you're really saying is $P=kV$ where $k=1$ atm/L. Let's keep track of this constant as the calculation progresses.
$$W=-\int_{1}^3P(V)dV=-\int_1^3kVdV=-\frac{1}{2}kV^2\bigg\vert_{1}^3\\=-\frac{1}{2}(1\text{ atm/L})(8\text{ L}^2)=-4\text{ atm}\cdot\text{L}$$
Now it's clear that you need to convert from atm$\cdot$L to J. Based on your information,
$$1\text{ atm}\cdot\text{L}=(1\text{ atm})(1\text{ L})=(1.013\times 10^5\text{ N/m}^2)(.001\text{ m}^3)=101.3\text{ N}\cdot\text{m}=101.3\text{ J}$$
which gives you your correct answer.

