Calculations on an irreversible adiabatic expansion


> 
> A $\pu{5.00 L}$ sample of $\ce{CO\_2}$ at $800\ \mathrm{kPa}$ underwent a one-step irreversible adiabatic expansion against a constant external pressure of $\pu{100 kPa}$. The initial temperature of the gas was $300\ \mathrm{K}$
> 
> 
> An alternative path between the initial and final states consists of a reversible isothermal expansion
>  from $5.00\ \mathrm{L}$ to the final volume $V$, followed by (reversible) constant volume cooling to the final temperature $T$.
> 
> 
> a) Give the equations for $\Delta U$, $q$ and $w$ for all three processes in terms of $V$ and $T$
> 
> 
> b) Calculate the final volume and temperature of the gas
> 
> 
> 


**My Attempt**


**Adiabatic Expansion**


$q = 0$ since it is adiabatic. $w = -100 \times (V-5)$. $\Delta U = q + w = -100 \times (V-5)$


**Isothermal Expansion**


$\Delta U = 0$. $w = -q = -\mathrm{nRT}\times \ln\frac V5$ = $-4000 \times \ln\frac{V}{5}$


$q = 4000 \times \ln\frac{V}{5}$


**Reversible Cooling**


$w = 0$. $\Delta U = \mathrm{c\_v \Delta T = c\_v \times (T -3oo)}$


$q = \mathrm{c\_v \Delta T = c\_v \times (T -3oo)}$


**Part b)**


$-100 \times (V-5) = \mathrm{c\_v \Delta T = c\_v \times (T -3oo)}$ since $\Delta U$ is a state function


$\mathrm{V = \frac{nRT}{p}} = 0.133T$ Now I am not sure if this step is correct. I used that $p = \pu{100 kPa}$. Is that right? From there I can solve for $T$ and hence for $V$.


Could someone please check if above working out is correct?


