what would the volume be at 8 degrees assuming constant pressure
2007-03-21 04:27:44 · 7 answers · asked by CoNfUsEd 2 in Science & Mathematics ➔ Chemistry
pV = nRT if p, n, are constant T V = kT
V1/V2 =T1/T2
V2 = V1T2/T1 T1: 24 =273+24 =297 T2 = 8+273 =281
V2 = 50 *281/293 =47.95 dm^3
2007-03-21 04:34:37 · answer #1 · answered by maussy 7 · 0⤊ 0⤋
The Gay - Lussac Law states that the volume is proportional to the absolute temperature (in Kelvins) at constant pressure. The relation between degrees Celsius and Kelvin is
T = t + 273.15
so
V = 50 dm^3 x 281.15 K / 297.15 K = 47.3 dm^3
Note that if you used degrees Celsius you would get an incorrect result!
2007-03-21 11:35:13 · answer #2 · answered by Bushido The WaY of DA WaRRiOr 2 · 0⤊ 0⤋
This is Charles' Law: "The volume of a gas is directly proportion to its volume at constant pressure".
Temperature must be in K
24°C = 297K: 8°C = 281K
T1V2 = T2V1
297 x V2 = 281 x 50
V2 = 281 x 50 ÷ 297
V2 = 47.31 dm³
2007-03-21 11:45:40 · answer #3 · answered by Norrie 7 · 0⤊ 0⤋
P1V1/T1 = P2T2/T2
P1 = P2 and can be igonored.
V1 / T1 = V2 / T2
v2 = V1 T2 / T1 = 50 dm^3 x 8 / 24
V2 = 16 2/3 dm^3 (or 16.666667 dm^3)
2007-03-21 11:36:04 · answer #4 · answered by Doctor Q 6 · 0⤊ 0⤋
We can use the Charles's Law:
V1 / T1 = V2 / T2
V2 = V1*T2 / T1 = (50)(8) / 24 = 16.66 dm³
That's it!
Good luck!
2007-03-21 11:35:10 · answer #5 · answered by CHESSLARUS 7 · 0⤊ 0⤋
assuming idea gas behavior "PV=nRT"
(V1/V2)=(T1/T2)
the volume at 8 degrees will be
V2=V1x(T2/T1)
= 50 dm^3 x (8+273)/(24+273)
= 47.3 dm^3
2007-03-21 11:40:14 · answer #6 · answered by Hussam a 1 · 0⤊ 0⤋
fun
2007-03-21 11:35:44 · answer #7 · answered by sweet m 1 · 0⤊ 0⤋