# Calculating Final Temperature and Volume of Adiabatic Expansion

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.

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

## 1 Answer

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.