Chemistry
posted by Liz .
A calorimeter is filled with 10.0 millimole (mmole) of methane gas and an excess of oxygen. When burned, the ignition wire releases 140.77 J of heat. The heat capacity of the calorimeter is 4.319 KJ°C1. qwire= 140.77 J
Calculate ΔHcomb. Recall that ideal gas law is PV=nRT where R=0.08206 Latm/molK (since you are working with atm not kPa) and the temperature is in Kelvin
Ccal= 4.319 KJ°C1
Initial temp= 21.1 degrees celsius
Final temp= 23.4 degrees celsius
v= 271 mL
P= 1.000 atm
I calculated delta H and got 979 KJ. I'm also not sure why ideal gas law formula would be used

I don't know the context of the problem but your 979 kJ/mol appears to be right if you have the number of significant figures correct. Does the problem say anything about "about 10.0 millimoles" methane? You have P, V, R, and T and you can calculate n. If I can calculate n I get something like 0.0112 and if the q is divided by 0.0112 instead of 0.01 to obtain per mol the number you get is much closer to 890 kJ/mol than if you use 0.01. Perhaps that 10.0 mmols is "close" and you are to use PV = nRT to solve for the real number of mmols. Here is the site I talked about a couple days ago where this experiment is delineated. There is nothing in the material about using PV = nRT BUT the instruction say to add a known amount of methane and I don't know how you would do that without using PV = nRT. I suspect you are to use PV = nRT to solve for the real number of mols CH4.
http://www.chm.davidson.edu/vce/calorimetry/heatofcombustionofmethane.html 
Ok thanks! I used PV=nRT to calculate the mols but I got a larger number. Here's what I did:
P= 1.000 atm
V= 0.271 L
R= 0.08206 Latm/molK
T= 2.3 K
and I got 1.436 as my answer 
NOPE. T is not 2.3, that's delta T for the experiment. The initial T is 23.1 so that must be the T of the methane when it's added to the bomb or 294.25K and that gives n = 0.0112 or about 11.2 millimoles.

ok I got it now!