Originally Posted by Bigron1 To answer your other question about a second part it also asks to evaluate the same for Boyles law. How do you only use one to get the answer for work done and thermal transfer? I thought you needed to combine them? 
Boyle's law states that there is a relationship between pressure and volume when the temperature is held constant.
$\displaystyle P \propto \frac{1}{V}$
Therefore
$\displaystyle P_1 V_1 = P_2 V_2$
Our previous calculation for pressure uses $\displaystyle T_2$ and $\displaystyle V_2$, so if we say that the pressure there is $\displaystyle P_2$, we can calculate $\displaystyle P_1$
$\displaystyle P_1 = \frac{P_2 V_2}{V_1} = \frac{895905 \times 0.4}{0.3} = 1194540 $Pa.
The thermal energy exchange is zero because the temperature doesn't change. We used the value T = 250 K.
The work done can be calculated using integration:
$\displaystyle W = \int P dV$
$\displaystyle = \frac{mRT}{M} \int_{0.3}^{0.4} \frac{1}{V} dV$
$\displaystyle = \frac{mRT}{M} \left[ \ln V \right]_{0.3}^{0.4}$
$\displaystyle = \frac{mRT}{M} \left( \ln 0.4  \ln 0.3 \right)$
$\displaystyle = \frac{mRT}{M} \frac{\ln 0.4}{\ln 0.3}$
$\displaystyle = \frac{5 \times 8.314 \times 250}{0.029} \frac{\ln 0.4}{\ln 0.3}$
$\displaystyle =  272733.6$ J
Blimey you answer posts late at night or early morning depending which way you look at it. zzzz
Thanks for the help

I am based on the UK, so the difference in timezone probably makes my post appear as if they are at unusual hours. However, I can assure you I'm not up in the middle of the night solving physics problems!