# One "mol" of an ideal gas, initially at "300 K", is cooled at constant "V" so that "P"_f is 1/4 "P"_i. Then the gas expands at constant "P" until it reaches "T"_i again. What is the work w done on the gas?

Dec 30, 2015

DISCLAIMER: Yes, this will be long!

This is one of those problems where drawing a "P-V diagram" would really help. Let's see approximately what that looks like for this problem.

We know that:

STEP 1
${T}_{i} = \text{300 K}$

${T}_{i} \downarrow \implies {T}_{f}$ such that:
${P}_{f} \downarrow \implies {P}_{i} \text{/} 4$
$\Delta V = 0$

STEP 2
${T}_{i} = \text{previous } {T}_{f}$
${V}_{i} = \text{previous } {V}_{i}$

${V}_{i} \uparrow \implies {V}_{f}$ such that:
${T}_{f} \uparrow \implies {T}_{i} = \text{300 K}$
$\Delta P = 0$

From that, we construct the PV-diagram as follows:

Notice how we have the possibility for a reversible work process that accomplishes the same thing more efficiently.

TWO-STEP PATH: STEP 1

We should know that inexact differential work $\partial w = - P \mathrm{dV}$ when expanding or compressing a gas, and that the inexact differential heat flow $\partial q$ is not $0$.

STEP 1: WORK

We can then proceed by noting that $\Delta V = 0$. Thus, any expansion/compression work is $0$:

$\textcolor{b l u e}{{w}_{1}} = - \int P \mathrm{dV} = \textcolor{b l u e}{0}$

STEP 1: HEAT FLOW

Since PV-work didn't happen here, it was all heat flow, conducting out through the surfaces of the container, slowly leading the gas particles to redistribute more loosely without changing in volume, thus naturally cooling the gas.

In other words, the change in internal energy was all due to heat flow at a constant volume ${q}_{V}$:

color(blue)(DeltaU = q_V = ???)

Unfortunately we don't know enough to determine ${q}_{V}$. Had we known whether the gas was monatomic, diatomic linear, or what have you, we could determine its degrees of freedom and write ${C}_{V}$ or ${C}_{p}$ in terms of the universal gas constant $R$, since basically:

\mathbf(DeltaU = q_V

$\setminus m a t h b f \left(= {\int}_{{T}_{1}}^{{T}_{2}} {C}_{v} \mathrm{dT}\right)$

Either way, this release of thermal energy means that the first work step was inefficient; thermal energy escaped the container while the work ${w}_{1}$ was being done.

TWO-STEP PATH: STEP 2

Okay, so what's ${w}_{2}$ then?

STEP 2: WORK

${w}_{2} = - \int P \mathrm{dV}$

$P$ is constant here, but the volume DOES change, so:

$\textcolor{g r e e n}{{w}_{2}} = - P {\int}_{{V}_{i}}^{{V}_{f}} \mathrm{dV}$

$= - P \left({V}_{f} - {V}_{i}\right)$

We don't know how the volume changed, but using the ideal gas law, we can fidget with this:

$= - \cancel{P} \left(\frac{n R {T}_{f}}{\cancel{P}} - \frac{n R {T}_{i}}{\cancel{P}}\right)$

$= \textcolor{g r e e n}{- n R \left({T}_{f} - {T}_{i}\right)}$

STEP 2: TEMPERATURE

Alright, so we know ${T}_{f} = \text{300 K}$ in step 2. What's ${T}_{i}$ in step 2? Well, we can use what we knew from step 1. With a constant volume in step 1:

$P V = n R T$

${P}_{i} {\cancel{{V}_{i}}}^{\text{constant") = cancel(nR)^("constant}} {T}_{i}$

${P}_{f} {\cancel{{V}_{i}}}^{\text{constant") = cancel(nR)^("constant}} {T}_{f}$

${P}_{i} / {T}_{i} = {P}_{f} / {T}_{f}$

The temperature from step 1 we want is ${T}_{f}$, which is ${T}_{i}$ in step 2, so:

${T}_{f} = \frac{{P}_{f}}{{P}_{i}} \times {T}_{i} = \frac{1}{4} \times 300 = \textcolor{g r e e n}{\text{75 K}}$

Thus, in step 2, the gas heated from ${T}_{i} = \text{75 K}$ to ${T}_{f} = \text{300 K}$. Finally, we're getting a value! So, with $R = \text{8.314472 J/mol"cdot"K}$:

$\textcolor{b l u e}{{w}_{2} = - \text{1870.76 J}}$

In expansion work, work is done by the gas, so it should be negative.

REVERSIBLE WORK PATH

Now, if we try that reversible path, we assume that steps 1 and 2 aren't there and that this is an efficient path where the gas is acted upon very slowly so that it has time to adjust. We know that in this path, $\Delta T = 0$, but $P$ and $V$ change. So:

REVERSIBLE WORK PATH: WORK

${w}_{\text{rev}} = - \int P \mathrm{dV}$

$= - \int \frac{n R T}{V} \mathrm{dV}$

$T$ is actually constant this time, so...

$\textcolor{g r e e n}{{w}_{\text{rev}}} = - n R T {\int}_{{V}_{i}}^{{V}_{f}} \frac{1}{V} \mathrm{dV}$

$= \textcolor{g r e e n}{- n R T \ln | \frac{{V}_{f}}{{V}_{i}} |}$

Notice how the negative natural logarithm approximates the shape of the reversible path.

REVERSIBLE WORK PATH: VOLUME

Okay, but how did the volume change? Inversely proportionally to the pressure! The volume increased at a constant temperature, so the pressure should decrease.

Using the same trick with the ideal gas law as before:

$\textcolor{g r e e n}{{w}_{\text{rev}}} = - n R T \ln | \frac{\frac{\cancel{n R T}}{P} _ f}{\frac{\cancel{n R T}}{P} _ i} |$

$= - n R T \ln | \frac{{P}_{i}}{{P}_{f}} |$

$= - n R T \ln | \frac{{P}_{i}}{\frac{1}{4} {P}_{i}} |$

$= \textcolor{g r e e n}{- n R T \ln 4}$

So, with $R = \text{8.314472 J/mol"cdot"K}$:

$\textcolor{b l u e}{{w}_{\text{rev" = -"3457.89 J}}}$

In expansion work, work is done by the gas, so it should be negative.