Question

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?

Answer 1

`V_1 = "13.2 L"`
`V_2 = "6.6 L"`

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This is asking you about the definition of work.

`\mathbf(w = -PDeltaV`

(`DeltaV = V_2 - V_1`.)

• You have a constant pressure, `P = "120 kPa"`.
• You also know from the question that `V_2 = 1/2V_1`.
• You know that `w = "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.

`-w/P = V_2 - V_1 = 1/2V_1 - V_1`

`w/P = V_1 - 1/2V_1 = 1/2V_1`

But since we're solving for `V_1` anyways...

`color(green)(V_1) = (2w)/P`

`= (2("790 J"))/("120 kPa") = color(green)(("1580 J")/("120 kPa"))`

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 `"1 J"` = `"1 L"cdot"kPa"`.

Consider the universal gas constant `R`:

`R = "8.314472 J/mol"cdot"K" = "0.083145 L"cdot"bar/mol"cdot"K"`

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

It would also help to know the conversion factor from `"Pa"` to `"bar"` to get the units to cancel:

`"1 bar" = 10^5` `"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, `\mathbf(R)` can be a conversion factor!

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

`= color(green)("15.80 L"cdot"bar")`

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

`color(blue)(V_1) = ("15.80 L"cdotcancel"bar")/(1.2 cancel"bar") = color(blue)("13.2 L")`

And `V_2 = "6.6 L"` because why not.