Expansion work by a gas when pressure is not explicitly constant?
You can calculate the work done and Heat exchange if you assume that the process is quassi-statically performed and it is reversible and most importantly done in a single step. 
 

Assume that the process through which the $\ce{He}$ is expanded obeys the equation, $PV^x = K$(constant). This equation can also be alternatively stated as, $TV^{x-1} = K$. Now, suppose the gas is transformed to the state $(T_2,V_2)$ from the state $(T_1,V_1)$. So, we can thus write, $$T_2V_2^{x-1} = T_1V_1^{x-1}$$ so, putting the given conditions, we can find $x$ as, $$x= 1+ \frac{\log\left(\frac{T_2}{T_1}\right)}{\log\left(\frac{V_1}{V_2}\right)} \approx 0.676$$ Thus, after knowing $x$ ,you can calculate the work done by the gas as,  $$W= \int_{V_1}^{V_2} P\mathrm{d}V = \int_{V_1}^{V_2} \frac{K}{V^x}\mathrm{d}V = \frac{P_2V_2 - P_1V_1}{1-x} = \frac{R(T_2-T_1)}{1-x}$$ We know all the value of $T_1, T_2 $ &amp; $x$, and work done can be easily calculated. and we already know the change in Internal energy  and thus heat exchange can be easily calculated and also $\Delta H $ can be calculated   ($Q = (C_V + \frac{R}{1-x}) \Delta T$ and $\Delta H = C_P \Delta T $) and thus all the quantitites can be known.