Work done in reversible isothermal expansion
I agree with getafix, if you would like an answer that is more tailored to you, you should show us exactly what you've done.

However, I am going to make a (hopefully educated) guess that what you did was to pull $p_\mathrm{ext}$ out of the integral. That is incorrect, because $p_\mathrm{ext}$ is not a constant here.

This process is known as an isothermal expansion - isothermal because the temperature remains constant throughout - and expansion because volume is increased. In thermodynamics it is very important to note which variables are held constant, because then that lets you decide which formula is appropriate to use, or how to derive such formulae).

Since the process is reversible, the external pressure must always be equal to the pressure exerted by the gas, which can be calculated via the ideal gas law $pV = nRT$. Therefore, you have (where 1 and 2 denote the initial and final state respectively)

$$\begin{align}
w &amp;= -\int_1^2 p\,\mathrm{d}V \\
&amp;= -\int_1^2 \frac{nRT}{V}\,\mathrm{d}V
\end{align}$$

and now since $T$ is a constant, you can take it out of the integral (along with $n$ and $R$ which are also constants)

$$\begin{align}
w = &amp;= -nRT\int_1^2 \frac{1}{V}\,\mathrm{d}V \\
&amp;= -nRT\ln\left(\frac{V_2}{V_1}\right)
\end{align}$$

Since you weren't given numerical values for $V_2$ and $V_1$, you can't evaluate a numerical value for $w$. The closest you can get is therefore

$$w = (-24.9~\mathrm{kJ})\ln\left(\frac{V_2}{V_1}\right)$$