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?