Calculating Final Temperature and Volume of Adiabatic Expansion

Question

A $\mathrm{5.00\ L}$ sample of $\ce{CO2}$ at $800 \ \mathrm{kPa}$ underwent a one-step (irreversible) adiabatic expansion against a constant external pressure of $100\ \mathrm{kPa}$. 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 equations (in terms of $V$ and $T$, the final volume and temperature of the gas) for $\Delta U$, $Q$ and $W$ for all three processes.

b) Briefly state why $\Delta U$ is the same for both paths.

c) Hence or otherwise calculate the final volume and temperature of the gas.

My Attempt

Part A

I can do this part. I am pretty sure I got the correct answers except for $\Delta U$ for the cooling process. Could you please check if my answers are correct.

Adiabatic expansion: $\Delta U = 33.33(T-300)$, $Q = 0$ and $W = 33.33(T-300)$

Isothermal expansion: $\Delta U = 0$, $Q = 4000\ \ln\frac{V}{5}$ and $W = -4000\ \ln\frac{V}{5}$

Cooling: $\Delta U = 33.33(T-300)$, $Q = 33.33(T-300)$ and $W = 0$

Part B

That is just because internal energy is a state function, independent of the path taken.

Part C

What confuses me here is the word 'hence', which implies that I need to use the fact that internal energy are equal for the two process. However I have no idea how to use this to find the final temperature and volume. I suspect that I might gotten the expression for $\Delta U$ wrong for the cooling process.

Answer 1

The final volume V and final temperature T can be determined exclusively from the information for the irreversible path if we assume that the gas continues to expand until it equilibrates at a final uniform pressure of $P=P_{ext}=100$kPa and a final uniform temperature of T. Let $T_0$ be the initial temperature (300 K), $P_0$ be the initial pressure (800 kPa), and $V_0$ be the initial volume (5 L). Then the number of moles is $$n=\frac{P_0V_0}{RT_0}$$ The work done on the surroundingsis given by:$$W=P_{ext}(V-V_0)$$ The change in internal energy is $$\Delta U=nC_v(T-T_0)$$ So, from the first law,$$nC_v(T-T_0)=-P_{ext}(V-V_0)$$ The final volume is given by: $$V=\frac{nRT}{P_{ext}}$$ These equations are sufficient to solve uniquely for the final temperature T and the final volume V exclusively in terms of the input data values.