A 2L lab sample is at 72degrees celcius a pressure of 4atm.?
Answer:
I just learned this yesterday in chem class... tho not 100% sure, I think I would use the following equation to solve it: [(P1xV2)/T1]=[(P2xV2)/T2], then solve for P2, so P2=(P1xV1xT2)/(T1xV2).
just plug in the constants:
V1=2L
T1=72deg C (i think u must convert to kelvin scale) so then = 72+273=345deg K
P1=4atm
T2=89deg C ----> (89+273)= 362deg K
V2=1.5L
the answer i got that way was: 5.6 atm , which kinda makes sense because the pressure should be higher due to the increase in temperature and drop in volume. I hope it makes sense, my 1st answer ever in the site, lol
I presume this sample is a gas, but it doesn't say.
Use the equation; P1V1/T1 = P2V2/T2.
(remember T has to be in Kelvin!)
4 x 2/[273+72] = P2 x 1.5/[273+89]
Solve for P2: [362 x 4 x 2/345]/1.5 = ~ 5.6atm
Confused by all those PVT formulas? Remember just 1: PV = nRT. R is always constant, and in this problem, no sample is lost or gained, so n is constant. So we can write-
PV/T = a constant = PV/T
4 x 2 / (273+72) = P x 1.5 / (273+89)
P = 5.6 atm
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