The volume of a gas is 4.30 L, measured at 1.00 atm. What is the pressure of the gas in mmHg if the volume is changed to 9.89 L?
2007-12-09 10:38:14 · 3 answers · asked by dmastrovich 1 in Science & Mathematics ➔ Chemistry
Use the formula P1V1 = P2V2:
solving for P2: P2 = P1V1/V2:
P2 = (1.00atm x 4.30L)/9.89L = 0.435atm:
Converting to mmHg:
0.435atm x 760mmHg/atm = 331 mmHg
2007-12-09 10:56:35 · answer #1 · answered by Flying Dragon 7 · 0⤊ 0⤋
5...Boyle's regulation. P1 x V1 = P2 x V2 a million.fifty 5 ÷ 5 = 0.31 atm. 6....Charle's regulation. V1 x T2 = V2 x T1 V2 = (a million.fifty 5 x 173) ÷ 3 hundred = 0.89 L 7...gay Lussac's regulation. P1 x T2 = P2 x T1 80 5.5 x 4 hundred = P2 x 3 hundred P2 = 34,two hundred ÷ 3 hundred = 114 kPa 8....Charle's regulation. T1 x V2 = T2 x V1 273 x 325 = T2 x 275 T2 = 88,725 ÷ 275 = 322.6 ok 9....Boyle's regulation. P1 x V1 = P2 x V2 V2 = 15 x 3 hundred ÷ 0.9 = 4,500 ÷ 0.9 = 5,000 mL dispersed to environment.
2016-12-17 12:43:20 · answer #2 · answered by Anonymous · 0⤊ 0⤋
P1V1 =P2V2
Solve for P2
P2 = P1V1/V2 = 760mm x 4.30L/9.89L = 330 mm Hg
2007-12-09 10:55:49 · answer #3 · answered by papastolte 6 · 0⤊ 0⤋