6
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ That 33.33 is low. The constant volume molar heat capacity of carbon dioxide at these temperatures is 3.47R. $\endgroup$ – Chet Miller Jan 5 '16 at 22:07
4
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.