How to compute the work Done in an adiabatic irreversible expansion?

Question

The question is as follows

What is the magnitude of work performed by one mole of an ideal gas when its volume increases eight times in an irreversible adiabatic expansion if initial temperature of gas is $\pu{300K}$.
$C_V$ for the gas is $\pu{1.5 }R$. ($R=\pu{2 Cal/mol/K})$
(A)$~\pu{900 Cal~}$ (B) $~\pu{450 Cal~}$ (C) $~\pu{675Cal~}$ (D)$~\pu{331.58Cal}~$

I feel like I do not have enough information. Am I missing something? I tried to use the relation $PV^\gamma$=$k$ ,But did not get much ahead.

I used $\frac {P_1 V_1}{T_1}$=$\frac {P_2 V_2}{T_2}$

i.e keeping the moles constant to find a relation between pressures and temperatures. But this still leaves me hanging as I cannot seem to get rid of variables.

Any more suggestions?

Here is the question directly from source

Answer 1

Assume the work is against constant pressure and that the final pressure is that which is worked against.

Then

$$\begin{align} w&=\Delta U = C_V \Delta T= C_V (T_f-T_i) \\ &= -p_\textrm{ext} \Delta V =-p_f(V_f-V_i) \\ &= -\frac{nRT_f}{V_f}(8V_i-V_i)\\ &= -\frac{7RT_fV_i}{8V_i}\\ &= -\frac{7RT_f}{8} \end{align}$$

Solving for $T_f$ and inserting $C_V=\frac32 R$ gives

$$T_f=\frac{C_VT_i}{\frac78 R + C_V}=\frac{1.5R T_i}{\frac78 R + 1.5 R}=\frac{1.5 T_i}{\frac78 + 1.5 }=\frac{12}{19} T_i = \pu{189.47 K}$$

Using the approximation $\pu{R= 2 cal mol-1 K-1}$ results finally in $$w = C_V(T_f-T_i)= \pu{ 3 cal mol-1 K-1}\cdot(\pu{189.47 K}-\pu{300.00 K})=\pu{-331.58 cal}$$