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?

#V_1 = "13.2 L"#
#V_2 = "6.6 L"#

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.