How to calculate new air pressure with temperature change? 
The question:  The gauge pressure in your car tires is $2.60 * 10^5 N/m^2$ at a temperature of $35.0°C$ when you drive it onto a ferryboat to Alaska. What is the gauge pressure later, when the temperature has dropped to −28°C? Assume that the volume has not changed. Answer needed in atm. 

Okay, I've been working on this problem for about two hours already and I just cant seem to make it work, I know I'm missing something with the equations but I cant seem to figure it out. I know that $PV=NRT$ and that in the problem I am given initial P, and T and R is the gas constant, leaving N and V as unknown. I also know that I can use V=RT/P to find the volume (I got .009886, and I don't think that is right) and then plug that back in to calculate N (I got 1.659795*10^-32, not sure about that) at the given temperature (35C=309.15K) but when I do all this I get an outrageous answer that cant possibly be right (P=5.703*10^-59? no way on earth). I don't know if I am maybe using the wrong formula, maybe the wrong constant, messing up on algebra somewhere or if my calculator is just fudging my answer (I do that sometimes). Someone please give me some pointers on how to set this problem up to get the right P at the end! 
Thank you so much for the help guys, I think I actually understand what is going on with the algebraic organization (for once). When I do it myself I get an answer of 207,016 Pa which is 2.0431 atm, but the automated system I am using keeps telling me that is a wrong answer. 
 A: You are using the equation $V=RT/P$ but you seem to have left out $N$. This is likely (one reason) why your answer doesn't make sense.
Since you are working with a tire that is presumable sealed, you know $N$ is constant. You are also told $V$ is constant. So consider $PV=NRT$. You can rearrange to get $N/V=\frac{P}{RT}.$ Since the LHS of the equation doesn't change, the RHS must not change either. So now you have a quantity $\frac{P}{RT}$ which is the same before and after. Sounds useful!
My thought process above is that I'm trying to find some single quantity that's the same before & after. Since both $V$ and $N$ are the same, why not combine them into something else? That quantity happens to be density, but that's not crucial here; just good to notice.
That might get you started. You may have to thing deeply about gauge vs absolute pressure. I know I certainly would.
A: Let's look at the terms of $PV=nRT$ before and after:
     Before     After
P    2.6x10^5   P2
V    V          V     (the question says "Assume that the volume has not changed")
n    n          n     (the number of molecules of gas is unchanged - 
                          nothing has got out or in)         
R    R          R     (universal constant)
T(C) 35C       -28C
T(K) 308.15K   245.15K 

First thing to note is that I have immediately calculated the temperatures in Kelvin. Centigrade/Celsius is just an easy way to get confused here - gas equations are all in Kelvin. Kelvin Kelvin Kelvin. Gas in Kelvin, Kelvin has gas. Commit to memory. (Sorry Kelvin).
So we know P, R and T(K) both before and after, and V and n (and R) are unchanged because there are no leaks and the question tells us so. Let us collect all the constants, then:
$$ PV = nRT \\
\frac{PV}{T} = nR \\
\frac{P}{T} = \frac{nR}{V} $$
Now everything on the right is constant (unchanged) from before to after, so the result of the left is unchanged. Give them names $1$ for before and $2$ for after:
$$\frac{P_1}{T_1} = \frac{nR}{V} \\
\frac{P_2}{T_2} = \frac{nR}{V}$$
Equate those constants (substitute):
$$\frac{P_1}{T_1} = \frac{P_2}{T_2} $$
Rearrange for our 1 unknown ($P_2$):
$$P_2 = \frac{P_1 T_2}{T_1}$$
Thus we have our unknown in terms of only knowns, and we can substitute and calculate like a good Engineer:
$$P_2 = \frac{2.6\times10^5 \times 245.15}{308.15} $$
(Kelvin Kelvin Kelvin)
$$P_2 = 2.07\times10^5 \text{Pa}$$
A: For isochoric processes $$\frac{p}{T}=const$$
which means the answer for your question:
$$p_2=\frac{p_1*T_2}{T_1}$$ where all temperatures are in Kelvin. For more information see http://de.wikipedia.org/wiki/Isochore_Zustands%C3%A4nderung (unfortunatelly english version is not so usefull this time)
