a sample of nitrogen gas occupies 1.55L at 27 degrees C and 1.00atm pressure. what will the volume be at -100degreesC and the same pressure?
2007-05-31 17:13:30 · 3 answers · asked by Anonymous in Science & Mathematics ➔ Chemistry
PV/T (before) = PV/T (after)
Temperature here have to be in Kelvin: which is just Celsius + 273
(1)(1.55)/(27 + 273 = 300) = (1)(X)/(-100 + 273 = 173)
Solve for X and you'll get the volume after the temp. have changed.
2007-05-31 17:38:21 · answer #1 · answered by meaepistula 1 · 0⤊ 0⤋
You use the combined gas law: pv/t=p[2]v[2]/t[2]. So p is 1.00 atm, v is 1.55L, and t is 300 kelvin (always use kelvin in gases). p[2] is also 1.00 atm, v[2] is the unknown, and t[2] is 173 kelvin. So, [(1.00 atm)(1.55L)]/300 = [(1.00 atm)(v)]/173. You then simplify to 1.55L/300 = 1.00v/173. When you solve for v[2] you get v[2]= 173(1.55/300). V=0.8938333L or 0.894L using significant digits
2007-05-31 17:42:19 · answer #2 · answered by TaylorK 2 · 0⤊ 0⤋
For constant pressure, V = k T
At the reference state 1.55 L = k (300)
k = 1.55/300
At the new state: V = k (173)
Just plug in and solve.
2007-05-31 17:24:01 · answer #3 · answered by cattbarf 7 · 0⤊ 0⤋