Final Temperature of a monatomic gas

[patrickmoloney]

Homework Statement

The molar energy of a monatomic gas that obeys van der Waals' equation is given by
$$E=\frac{3}{2}RT - \frac{a}{V}$$

where $V$ is the molar volume at temperature $T$ and $a$ is a constant.
Initially, one mole of such a gas is at temperature $T_1$ and occupies a volume $V_1$. The gas is allowed to expand adiabatically into a vacuum, so that it occupies a total volume of $V_2$. What is the final temperature of the gas?

Homework Equations

The Attempt at a Solution

This is what I have, I'm not sure if it's correct.

$$\begin{align*}<br /> \Delta E & = \frac{3}{2}R\Delta T - \frac{a}{\Delta V} \\<br /> & = \frac{3}{2}R(T_f - T_i) - \frac{a}{(V_f - V_i)} \\<br /> & = \frac{3}{2}RT_f - \frac{3}{2}RT_i - \frac{a}{(V_f - V_i)}<br /> \end{align*}$$

re-arranging for $T_f$,

$$T_f= \frac{2}{3R}(\Delta E + \frac{a}{V_f - V_i})+T_i$$

 

[DrClaude]

If $A = 1/B$, then $\Delta A \neq 1/\Delta B$.

Also, from the problem statement, you know something about $\Delta E$.

 

[patrickmoloney]

If $A = 1/B$, then $\Delta A \neq 1/\Delta B$.

Also, from the problem statement, you know something about $\Delta E$.

The system is adiabatic so the there is no heat transfer. Meaning $\Delta E = 0$

 

[Chestermiller]

The system is adiabatic so the there is no heat transfer. Meaning $\Delta E = 0$

That is not sufficient to conclude that $\Delta E=0$

 

[patrickmoloney]

That is not sufficient to conclude that $\Delta E=0$

The system is adiabatically insulated, no heat flows into the system; $$Q=0$$ Furthermore, the system does no work in the process; $$W=0$$ Thus it follows by the first law of thermodynamics that the total energy of the system is conserved; $$\Delta E =0$$

 

[patrickmoloney]

What about this:

$$\begin{align*}<br /> \Delta E & = \frac{3}{2}RT - \frac{a}{V} = 0 \\<br /> & = C_{v}T - a \Big{(}\frac{1}{V_f}- \frac{1}{V_i}\Big{)}<br /> \end{align*}$$
and therefore
$$\Delta T = \frac{a}{C_v}\Big{(}\frac{1}{V_f}- \frac{1}{V_i}\Big{)}$$

and

$$T_f= \frac{a}{C_v}\Big{(}\frac{1}{V_f}- \frac{1}{V_i}\Big{)}+T_i$$