Thanks a lot for the help, LowlyPion! :D
I've got the answer..
.00542 moles of air enters the can
Your method did work..
P1V1=P2V2
V2 = ((82093)(0.005*0.15)) / (1x10^5)
V2 = 6.157x10^-4 m^3
ΔV = V(can) - V2
ΔV = (0.005*0.15) - 6.157*10-4
ΔV = 1.343x10^-4
n=PV/RT = ((100000)(1.343x10^-4)) /...
Oh ok i got that part now(ΔV). thanks.
so basically what that's asking is what change in volume will occur when the pressure increases from 82093 to 100,000, right?
I forgot to subtract the overall volume, not sure why.. i did on my first attempt.
Anyway, so now would I still use n = PV/RT ...
hmm not sure i understand.
do you mean:
P1V1=P2V2 with P1 = 82093 and P2 = 1atm?
V2 = ((82093)(0.005*0.15)) / (1x10^5)
V2 = 6.157x10^-4 m^3
then do the n = PV/RT with V = V2 = 6.157x10^-4 and P = 1atm and T =298 ?
n= 0.0248 moles
still doesn't seem right.. although i am sure i...
[SOLVED]AP Question involving Ideal Gas Law - Thermal Physics - Pressure
I am having some trouble with an old ap problem(PART C ONLY), see below:
Homework Statement
1996 AP PHYSICS B Free Response Question:
The inside of the cylindrical can has a cross sectional area of 0.005 m^3 and...
W=Fd
W=(mg)h
W=(rho)Vgh
Note: (rho) = density - in this case of water
W=1000*.35*9.8*85
W=291550 Joules
P=W/t
P=291550/7200
Note 2hrs = 7200s
P=40.5 watts
Next one you have done correctly..
A1v1=A2v2
V2 = 2.88 m/s
Last one.. You must use bernoullis principle
P+(.5(rho)v^2)+((rho)gh)
P is...