Calculating Isothermal Expansion Work - 2 Moles Ideal Gas, 380K

[BlackMamba]

Hello,

I have what should be an easy problem, but I, of course, am having some issues.

Here's the problem: The temperature of 2 moles if an ideal gas is 380 K. How much work does the gas do in expanding isothermally to 3 times its initial volume?

So I know I'm going to to need to use this equation to find the work done:
$W = nRT ln (\frac{V_f}{V_i})$

I know that:

n = 2
R = 8.31
T = 380
Vi = (3)Vi
Vf = ?

I'm hung up on the volume portion of this problem. I don't have an initial or final volume and that's what I can't seem to work around. Any help would be greatly appreciated.

Thanks in advance.

 

[tigger]

Vi = Vi
Vf = 3 Vi
since it expands three times its initial volume

 

[manesh]

please correct vi=3.vi to be vf=3 vi, so now the problem is solved..

 

[omgitsroy326]

W = nRTln(3)

(Vf/Vi ) = 3Vi/Vi = 3

 

[BlackMamba]

Oh Lord. Something so simple that I couldn't put together. Thank you everyone for your reply. It is greatly appreciated. :)

 

[nvpraneeth]

simple

Hello,

I have what should be an easy problem, but I, of course, am having some issues.

Here's the problem: The temperature of 2 moles if an ideal gas is 380 K. How much work does the gas do in expanding isothermally to 3 times its initial volume?

So I know I'm going to to need to use this equation to find the work done:
$W = nRT ln (\frac{V_f}{V_i})$

I know that:

n = 2
R = 8.31
T = 380
Vi = (3)Vi
Vf = ?

I'm hung up on the volume portion of this problem. I don't have an initial or final volume and that's what I can't seem to work around. Any help would be greatly appreciated.

Thanks in advance.

vfinal=3vinitial