Since the gas acts like an ideal one, we start with the equation of state for an ideal gas:
We also have the equation for the work done on or by an ideal gas during an isothermal volume change:
We solve for the number of moles n in the first equation and plug it into the second to get rid of that.
The addition of the initial subscrips to the pressure and volume from the equation of state for an Ideal Gas comes because we have the gas in its initial state which is known and we compress it to a state where we only know the volume and temperature.
We now create two equations of work, one for the work needed to compress the gas to one half of its originial volume and a second to find the work to compress to one tenth of its volume.
First the equation to compress to one half of its original volume:
The negative number is okay because it's the work that was done on the gas to compress it...
Now for the second equation:
Now we just divide the two to find how more work is needed to compress to one tenth compared to one half:
It takes 3.32 times more work to compress the air.