# An ideal gas is compressed at a constant pressure of 120 kPa to one-half of its initial volume. The work done on the gas is 790 J. What was the initial volume of the gas?

Feb 18, 2016

${V}_{1} = \text{13.2 L}$
${V}_{2} = \text{6.6 L}$

\mathbf(w = -PDeltaV

($\Delta V = {V}_{2} - {V}_{1}$.)

• You have a constant pressure, $P = \text{120 kPa}$.
• You also know from the question that ${V}_{2} = \frac{1}{2} {V}_{1}$.
• You know that $w = \text{790 J}$. It is numerically positive because you are compressing the gas, i.e. doing work on the gas.

Therefore, you have everything you need to substitute and solve.

$- \frac{w}{P} = {V}_{2} - {V}_{1} = \frac{1}{2} {V}_{1} - {V}_{1}$

$\frac{w}{P} = {V}_{1} - \frac{1}{2} {V}_{1} = \frac{1}{2} {V}_{1}$

But since we're solving for ${V}_{1}$ anyways...

$\textcolor{g r e e n}{{V}_{1}} = \frac{2 w}{P}$

$= \left(2 \left(\text{790 J"))/("120 kPa") = color(green)(("1580 J")/("120 kPa}\right)\right)$

Almost there, but wait a minute. So we have energy divided by pressure. We need volume. Let's just say that you don't know that $\text{1 J}$ = $\text{1 L"cdot"kPa}$.

Consider the universal gas constant $R$:

$R = \text{8.314472 J/mol"cdot"K" = "0.083145 L"cdot"bar/mol"cdot"K}$

Notice how we have just equated $\text{J}$ with $\text{L"cdot"bar}$, a $\text{P"cdot"V}$ unit. Neat. That's how we can get volume.

It would also help to know the conversion factor from $\text{Pa}$ to $\text{bar}$ to get the units to cancel:

$\text{1 bar} = {10}^{5}$ $\text{Pa}$

That's all we need to convert into units that will cancel out to give a volume. Let's convert the denominator and then the numerator.

120 cancel"kPa" xx (1000 cancel"Pa")/cancel"1 kPa" xx "1 bar"/(10^5 cancel"Pa") = color(green)("1.2 bar")

Now for the numerator. Here's something cool. That's right, $\setminus m a t h b f \left(R\right)$ can be a conversion factor!

1580 cancel"J" xx ("0.083145 L"cdot"bar")/(8.314472 cancel"J")

$= \textcolor{g r e e n}{\text{15.80 L"cdot"bar}}$

And finally, with the right units, we get the initial volume to be:

$\textcolor{b l u e}{{V}_{1}} = \left(\text{15.80 L"cdotcancel"bar")/(1.2 cancel"bar") = color(blue)("13.2 L}\right)$

And ${V}_{2} = \text{6.6 L}$ because why not.