anonymous
  • anonymous
A steel cylinder capacity 10L holds 20 Moles of O2 at 27 C an imperfection causes the gas to leak out slowly so that temperature remains Contant Ext Pressure is 1 atm The cylinder Finaaly Contain 1 atm Calculate Work Done in Calories
Physics
  • Stacey Warren - Expert brainly.com
Hey! We 've verified this expert answer for you, click below to unlock the details :)
SOLVED
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
katieb
  • katieb
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
experimentX
  • experimentX
this is isothermal process, work done is given by \[ n R T \log \left( V_b \over Va\right)\\ V_a = {10 \text{L}} \\ V_b = {300.4 \over 273.4}*20*22.4 \text{L}\]
anonymous
  • anonymous
|dw:1351149292211:dw|
anonymous
  • anonymous
|dw:1351149481464:dw|

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

anonymous
  • anonymous
|dw:1351149608767:dw|
anonymous
  • anonymous
|dw:1351149969466:dw|
anonymous
  • anonymous
|dw:1351150077826:dw|
anonymous
  • anonymous
I can't continue there is not Q ?
anonymous
  • anonymous
Pay attention that in this case n is variable(leaking). so U is not constant(Adiabatic) So you can not find W(work) by int Pdv.
Vincent-Lyon.Fr
  • Vincent-Lyon.Fr
Is the answer 48.9 kJ = 11.7 Cal ?
anonymous
  • anonymous
|dw:1351152257623:dw|
anonymous
  • anonymous
I think you should have how percent was leaked ?
anonymous
  • anonymous
it is ~11700 cal @Vincent-Lyon.Fr
anonymous
  • anonymous
@experimentX : \[W=2.303*n*R*T* \log\frac{ P1 }{ P2 }\]
anonymous
  • anonymous
\[P1 V = nRT\] \[P1=~48atm\]
anonymous
  • anonymous
P2=1atm (given)......Can t we Do like this...?
experimentX
  • experimentX
i guess you can.
anonymous
  • anonymous
But Not getting the Correcr Result:
anonymous
  • anonymous
but @Vincent-Lyon.Fr Got it correctly...)
anonymous
  • anonymous
@Kryten
experimentX
  • experimentX
http://www.physicsforums.com/archive/index.php/t-530754.html

Looking for something else?

Not the answer you are looking for? Search for more explanations.