# Question 668a7

Mar 20, 2015

Here's a nice alternative to converting the pressure to $\text{Pa}$ and the volume to ${\text{m}}^{3}$ in order to get the answer in Joules.

In your case, the expression of the work done by the system is

$W = - {P}_{\text{exterior}} \cdot \Delta V$

You can first calculate the work in $\text{L" * "atm}$ and then convert to $\text{Joules}$ by using two expressions of the ideal gas constant, $\text{R}$.

Since $\text{R}$ is both equal to

"R" = 8.314 "J"/("mol" * "K"), (1) and

"R" = 0.08206 ("atm" * "L")/("mol" * "K"), (2)

you can divide the two values and come up with a conversion factor that'll take you from atm L to Joules

$\frac{\left(1\right)}{\left(2\right)} \implies \left(8.314 \text{J"/(cancel("mol" * "K")))/(0.082("atm" * "L")/(cancel("mol" * "K"))) = "8.314 J"/("0.082 atm" * "L}\right)$

So, if you calculate the work done by the system this way you'll get

$\text{W" = -"4.1 atm" * (70.8-24.8) * 10^(-3)"L" = "-0.18856 atm" * "L}$

In Joules, this would be

-0.1886cancel("atm" * "L") * "8.314 J"/(0.08206cancel("atm" * "L")) = "-19.1 J"#

So, if you ever need a quick way to get to Joules from atm L, use the ratio of the two values for the ideal gas constant.