can at 20 degrees C is thrown into a fire and is heated to 600 degrees C, how many times atmospheric pressure is the new pressure inside the can?
2006-06-21 05:59:45 · 5 answers · asked by jlynn 1 in Science & Mathematics ➔ Physics
P/T=P/T
1/(20+273)=P/(600+273)
P=30 atm
The new pressure would be 30 atms, but this is considering that the volume remains constant.
2006-06-21 06:07:06 · answer #1 · answered by Tim 4 · 0⤊ 0⤋
The ideal gas law states that - PV = nRT
P - Pressure
V - Volume
n - number of mols
R - constant
T - temperature
First change each temperature to degrees of Kelvin (add 273 to degrees C).
You know that the temperature changed, R cannot change (it is a constant), n did not chnage (the number of molecules stayed the same, they just got excited), and V did not change (unless P gets so big that it make the can burst.
So you are left will only two variables that change (T and P), which are proportional, so P1 * T1 = P2 * T2
T1 = (20 +273) K
P1 = 1 atm
T2 = (600 + 273) K
2006-06-21 06:12:36 · answer #2 · answered by ? 3 · 0⤊ 0⤋
Using the reltion P/T = constant
1/(273+20) = P/(273+600)
or P = 873/293 = 2.979522 atm.
2006-06-21 06:04:54 · answer #3 · answered by ag_iitkgp 7 · 0⤊ 0⤋
PV=nRT
V,n,R constant
P1/T1 = P2/T2
P2 = (1)*(600+273)/(20+273)
2006-06-21 06:05:14 · answer #4 · answered by scott_d_webb 3 · 0⤊ 0⤋
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2006-06-21 06:03:59 · answer #5 · answered by Bharat G 1 · 0⤊ 0⤋