I'm looking for an equation (or already existing online calculator)
that will help me determine how much pressure must be exerted on a
parcel of air to get the interior of it to increase to a desired
temperature. Say I wanted the inside of a 10" X 10" parcel full of air
to increase 10C, how much pressure (psi) would I have to exert on it?

Go feel of your car's tires and tell me if they feel hotter than
ambient temp (assuming that the car has not been driven lately). An
increase in pressure does not mean an increase in temperature.
However, an increase in temp will cause a pressure increase.

maybe I was unclear in my question. I assumed that the pressure would
result in a reduction of parcel volume. This then leads to an increase
in temp inside it. The little I remember from High School physics had
something to do with an "adiabatic process."

Oh, you kids with your questions are so cute! 10" x 10" is only 2
dimensions, Sparky. You need one more to have a go at the problem.

Let's see, PV=nRT, if memory serves.

T is expressed in Kelvins
n is moles
P is atmospheres
V is litres.
R is the gas constant. For your use it would likely be 0.08206 as you
are using L atm/mol K. If you use pascal, joules, or calories, the
constant is different. You can look it up; there aren't a lot.

A bit of a snag for your question is the "air" thing. The ideal gas
law doesn't recognize "air." You must calculate for nitrogen, oxygen,
a variety of other lesser and traces gases, and then mush them all
together in a great big gloppy mess! My suggestion is to use nitrogen.

For PV = nRT, figuring the change in the system the way you want gives us
(PV)/T = nR (and nR is a constant in this situation because the moles
don't change and the constant doesn't change).

So you just need to solve:

(P1V1)/T1 = (P2V2)/T2 for when T2 is T1 + 10.

Good luck!

(Jeez, I hope I didn't leave anything out. It's been a lotta years
since I had to do an ideal gas equation!)

Hello.  Redhoss and Markvmd are barking up the wrong trees.  Adiabatic
(means no heat transferred) compression does cause an increase in
temperature.  The beginning and ending states of an adiabatic
reversible process are described by the relationship:

P1 * V1^gamma = P2 * V2^gamma

P1 and P2 are initial and final pressure.
V1 and V2 are initial and final volume.

gamma is the ratio of spefic heats, Cp/Cv.  
For an ideal diatomic gas, gamma = 1.4
At normal pressures and temperatures, air is very well approximated as
an ideal diatomic gas.  It is over 99% N2 and O2.

You haven't really provided enough information to fully answer the
question.  The additional pressure required to achieve a given
increase in temperature depends on the initial pressure and
temperature.  Also, your wording suggests you envision increasing
pressure while keeping the volume the same (10" by 10" or whatever). 
This would require the addition of more gas, and so the process would
not be adiabatic.  Let's assume you are going to increase the pressure
by decreasing the volume, keeping the container sealed.

In that case, from the relationship above, together with 
P1 V1 = n R T1    and    P2 V2 = n R T2, 

you get:  P2 = P1(T2/T1)^(gamma/(gamma-1)) 

If you start at atmospheric pressure (101000 Pa or 14.7 psi) and 20
degrees C (293 K), and end at 30 degrees C (303 K), that gives:

P2 = 14.7*(303/293)^3.5   =  16.5 psi

So you only have to increase the pressure by 1.8 psi to get a 10
degree increase when starting at room temperature and pressure.  (I
used psi instead of SI units because that's what you put in your
question.)

Perfect. That's exactly what I was looking for. Thanks, rracecarr!