Tuesday 26 February 2013

The partial pressure of ethane over a solution containing 6.56 × 10–3 g of ethane is 1 bar. If the solution contains 5.00 × 10–2 g of ethane, then what shall be the partial pressure of the gas

The partial pressure of ethane over a solution containing 6.56 × 10–3 g of ethane is 1 bar. If the solution contains 5.00 × 10–2 g of ethane, then what shall be the partial pressure of the gas?

Answer
Step -1 (What is given in problem ?)
condition -1
Partial pressure of ethane = 1 bar
Mass of ethane =6.56 × 10–3 g
Condition -2
Now mass of ethane is 5.00 × 10–2
We have to find the pressure

So using Henry’s law 
Mass of dissolved gas(m) of directly proportional to its pressure (p)
m= kp   here k is proportionality constant
so plug the values we get
m1= kp1 ……………(1)
m2 = kp2 …………...(2)
Divide equation (2) by (1) equation we get
M2/m1 = p2/p1
Multiply by p1 both sides we get
P2 = m2*p1/m1
Plug the values now we get
                           P2 = 5.00 × 10–2* 1/6.56 × 10–3 g
                         P2= 7.62 bar (Answer)
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5 comments:

  1. Unknown17 April 2017 at 11:00

    In henry law, p=kx then how it become m=pk?

    ReplyDelete
    Replies
    1. Hadiya Tahir21 April 2018 at 23:25

      Solubility of mass of gas is directly proportional to pressure by henry law so m=kp

      Delete
    2. Master17739 June 2019 at 17:46

      Let me clear all this folks for ya!!
      Mole Frac. = Mass of solute/Molar m of solute/Total moles =Wb/Mb / Wb/Mb+Wa/Ma
      You know how its written
      So we know here the mass of ehtnae is 6.56 × 10^(-3) and even dividing it by its molar mass will give a even smaller value so we neglect moles of ethane in total moles so now
      Xb=Mole frac=Wb/Mb / Wa/Ma
      Take Wa as x gm and substitute the value of Kh/x fron first equation by taking Wb as 6.56 × 10^(-3) in second equation and there you have it !! IF ANY doubt please reply !

      Delete
  2. Unknown16 May 2019 at 12:27

    Yeah ur absolutely correct

    ReplyDelete
  3. Master17739 June 2019 at 17:45

    Let me clear all this folks for ya!!
    Mole Frac. = Mass of solute/Molar m of solute/Total moles =Wb/Mb / Wb/Mb+Wa/Ma
    You know how its written
    So we know here the mass of ehtnae is 6.56 × 10^(-3) and even dividing it by its molar mass will give a even smaller value so we neglect moles of ethane in total moles so now
    Xb=Mole frac=Wb/Mb / Wa/Ma
    Take Wa as x gm and substitute the value of Kh/x fron first equation by taking Wb as 6.56 × 10^(-3) in second equation and there you have it !! IF ANY doubt please reply !!

    ReplyDelete

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