Hello, Ja,
A gas law problem that does not involve a change in the moles of that gas can generally be solved with the expression:
Where the two states are noted by the subscripts 1 and 2. This problem meets that definition. The expression is the result of two ideal gas law equations representing the before (1) and after (2) states, but is a lot simpler since both n, the moles, and R, the gas constant, cancel out.
All temperatures must be converted to Kelvin for the gas laws. Add 273.15 to Centigrade to reach Kelvin.
Rearrange the above equation so that we isolate the unknown, T2, in this case:
T2 = T1(V2/V1)(P2/P1)
Note how I've arranged the volume and pressure into ratios. It helps us visualize what's happening to the gas. Our volume is increasing, so the temperature must increase. Pressure stays the same, so (P2/P1) cancels out in this problem. Thus we will see the temperature increase. Does this make sense? Yes, because if we are to maintain the pressure while increasing the volume 4-fold, the temperature will have to rise by a similar(exact) amount.
PEnter the data and solve. I get 400.6K. Subtract 273.15 to get Centigrade, if you insist. That leaves us with 127.45C, or 127C with 3 sig figs.
Bob
