[simpleton]

Question:
1 mol of ideal gas undergoes an isothermal expansion at 300K from 1.00 atm to a final pressure while performing 200J of expansion work. Calculate the final pressure of the gas if the external pressure is 0.20atm.

My attempted answer:
As it is isothermal expansion of ideal gas it will be a reversible expansion. So i used this formular:
w = -nRT In(Pinitial/Pfinal)
-200J (since it is expansion) = - 1 (8.314) (300) x In (1/Pfinal)
0.08 = In 1/Pfinal
e 0.08 = 1/P final
1.08 = 1/P final
Hence P final = 1/1.08 = 0.92 atm

But the correct answer should be 0.71 atm. What have I done wrong here?

I really appreciate all the helps! and many thanks in advance! 

[MrTeo]

Remember that there's an external pressure of 0.20atm.

[simpleton]

Hmm yeah. but i don't know what to do with that. Is it because it's a wrong formular?

[MrTeo]

Nope, everything's right... the only thing to change is the starting pressure of the gas. If there's an external pressure of 0.20atm the pressure of the ideal gas is more than 1atm (which is the total pressure).

So we'll have p_gas=1.2atm and this pressure must be summed up with the final pressure too.
We'll have: exp(k)=(p_gas)/(p_x+0.20).

[simpleton]

oh man! i'm so dumb!  ughh!

THANKS A LOT! i guess i need to solve a lot more problems before i get my head clear with it. THANK YOU VERY MUCH! 

[Fridushka]

ohhh no dont say like that!!!..everyone needs to practice and everything need practice...never worry about it! good luck !