急!!!!!!!!!!唔清楚有關BOYLE LAW THXX

tze

2011-08-03 10:52 am

the piston pump sucks out a volume of 0.1 m^3 in each stroke
the vessel to be evacuated has a capacity of 1.5M^3 and the pressure inside in originallt at 1atmospehric pressure.the piston is pushed and pulled slowly to expel air from vessel,auusme Temperature is constant during pumping process

(B)after 5 stroke

答案話要這樣計:
pressure=1乘(1.5/1.6)^5

我唔明點解要乘5次方啊!!!!!!!!!!!!!!!!!(仲有點解有1.6存在?)THXXXXXXXXXXXXXXXXXX

更新1:

佢話要計AFTER FIVE STROKE 的PRESSURE 我唔明點解要乘5次方同1.6怎樣來

更新2:

點解個PREESURE會轉 GE~ 個VOLUME一直固定而TEMTEMPATURE無變好奇怪喔@@ !!!!!!!!! initial pressure = P1 atm initial volume = 1.5 m^3 initial pressure = P2 atm initial volume = 1.5 m^3 THX

更新3:

THXXXX.不過我想問點解P1 atm 同P2 atm都唔同 V應該唔同WOW~~ 即係兩個intial pressure (P1同P2)都唔同 (TEMPERATURE一樣) 個INITIAL VOLUME就相同咁樣咪P1*V=/P2*V thank u

更新4:

(短暫PRESSURE相同 WHY?)

回答 (1)

天同

2011-08-03 4:41 pm

✔ 最佳答案

After 1 stroke, 0.1 m^3 of air is pumped out. This is equivalent to expanding the air inside the vessel from a volume of 1.5 m^3 to (1.5+0.1) m^3 = 1.6 m^3

Applying Boyle's Law,
initial pressure = 1 atm
initial volume = 1.5 m^3
final pressure after 1 stroke = P1
final volume after 1 stroke = 1.6 m^2

hence, 1 x 1.5 = P1 x 1.6
i.e. P1 = 1.5/1.6 atm

After the 2nd stroke
initial pressure = P1 atm
initial volume = 1.5 m^3
final pressure after 2 strokes = P2
final volume after 2 strokes = 1.6 m^2
hence, P1 x 1.5 = P2 x 1.6
P2 = (1.5/1.6).P1 = (1.5/1.6).(1.5/1.6) atm = (1.5/1.6)^2 atm

After the 3rd stroke
initial pressure = P2 atm
initial volume = 1.5 m^3
final pressure after 3 strokes = P3
final volume after 3 strokes = 1.6 m^2
hence, P2 x 1.5 = P3 x 1.6
P3 = (1.5/1.6).P2 = (1.5/1.6).(1.5/1.6)^2 atm = (1.5/1.6)^3 atm

Therefore, it can be deduced aftr 5 strokes, the pressure P5 is
P5 = (1.5/1.6)^5 atm

2011-08-03 14:08:06 補充：
Q:點解個PREESURE會轉 GE~ 個VOLUME一直固定
When the piston is pulled up, the volume of the gas expands from 1.5 m^3 to 1.6 m^2. The pressure in the vessel is thus decreased.

2011-08-03 14:10:53 補充：
Q: 而TEMTEMPATURE無變好奇怪喔@@
This is given in the question. Such assumption is valid when the pumping is done slowly and the vessel is not insulated such that heat flow to and from the gas is possible.

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