[a confused chiral girl]

I am having trouble calculating this question: Calculate the work ,w,  when 1.2 litre of an ideal gas at an initial pressure of 3.4 atm is expanded or compressed isothermally to a final volume of 3.4 litres reversibly.

I used this formula that I know for work: w= -P (V) and I plugged in the numbers w = -3.4 (3.4-1.2), which gives me -757.911 Joules. However, the answer is supposed to be -430.5 J. PLease show me how to get that correct ansewr.
thanks!! 

[Yggdrasil]

In calculations in science you have to worry about UNITS.  You calculated work in units of L*atm.  To get to -430.5, you need to convert L*atm to J.

[a confused chiral girl]

I don't understand, because I thought I already converted L*atm to Joules. Before, calculating -3.4(3.4-1.2) gves me -7.28L*atm, so I used this number to multiply 101.325 in order to change units to Joules. This gave me -757.911 J. the right answerthough,  is -430.5....please show me how to calculate. thank u very much.

[enahs]

Your key to thermodynamic calculations.

Learn all your keywords, like in this case isothermal and reversible; and what they mean to a thermodynamic systems. In many calculations, just by knowing the keywords you can immediately recognize the answer is 0, which helps you greatly in calculating the other parts of the questions (but that is not the case here, just fyi).

In this case, the work is isothermal and reversible, the equation you are using is for constant pressure.
You must use your equation (or derive one, it is quite easy) for the isothermal expansion work.

[a confused chiral girl]

hi, so what you are trying to confirm is that the equation I have used is wrong. I was thinking of PV = nRT, but it does not contain the variable , w for Work. Please help me out here. I have been struggling with this problem for quite a while now, I am really confused. with what equation I need to use here.

[chiralic]

In your exercise, don't tell you that the pressure is CONSTANT...so you can't use the formula
w=P(Vf-Vo)...

[chiralic]

Hint: w= PdV => use PV=nRT => P=nRT/V then w=nRT(dV/V)  (here you need use integral calculus), You know V_inicial, V_final and P_inicial

[enahs]

Here is a quick google result for isothermal work:
http://hyperphysics.phy-astr.gsu.edu/hbase/thermo/isoth.html
Where you can see the equation, and maybe understand where it comes from.

Again, the most important thing you can do for your thermodynamics is learn then key words and conditions and what they imply; as it either tells you how to work the problem, or in many cases simplifies the problem a tremendous amount.

[damas17]

Although this post is quite old, I will post the answer ('{' means integral)
W = - {_vi^vf P(V).dV
W = - {_vi^vf(nRT/V).dV
W = - nRT{_vi^vf dV/V
W = - nRT[ln(V)]_vi^vf
(as nRT = Pi.Vi, initial pressure*initial volume)
W = -(Pi*Vi)*[ln(Vf)-ln(Vi)]
W = -(Pi*Vi)*ln(Vf/Vi)
------------------------- Calculation ----------------------
Pi = 3.4 atm = 3.4*101325 Pa = 344 505 Pa
Vi = 1.2 L = 0.0012 m³
ln(Vf/Vi) = ln(3.4/1.2) = 1.04145387 (It doesn't matter if You convert Vf and Vi to m³ or not, as long as they have the same unit.)
W = - 344505*0.0012*1.04145387 = -430.543279 J
-----------------------------------------------------------
PS : I've never studied thermodynamics ^^ hahaha