# 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

$q = 0$ 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?

• I didn't check you arithmetic, but your methodology is correct. – Chet Miller Feb 22 '16 at 11:56
• @ChesterMiller thanks, that is fine. Just to make sure, it is correct to use p as 100 kPa for finding the volume? – Nanoputian Feb 22 '16 at 12:04
• Yes. The implication of the problem statement is that you allow the gas to continue to expand until it equilibrates at its final state. – Chet Miller Feb 22 '16 at 12:10