Thermodynamics First law ideal gas question

[deeko1987]

Homework Statement

We have
0.0008 Kmol of an ideal gas are expanded from V1 to V2 v2=3V1
process is reversible and T/V=Constant

if the work obtained by this expansion is 9.4 KJ find the initial temperature

R=8.314 KJ Kmol-1

Homework Equations

PV=nRT

possibly T/V = T/V

The Attempt at a Solution

The method I thought to use is as follows.

PV=nRT we know work = 9.4KJ

9.4/nR = T2

T2/V2 = T1/V1

multiply T2/V2 by V1

and we get T2/3 (because the ratio is 1/3)

got 471.092 kelvin for T1

a class mate used this method

PV= 9.4KJ

Pressure is constant as t/v=constant

P(3V1-V1) = 9.4KJ

2PV1= 9.4 KJ (divide by 2)
PV1 = 4.8KJ

PV1=nRT2 = 4.8

solving for t

4.8/nR = 721

T1 = 721 kelvin

Could somebody please explain which method is correct to use and if it is the second method the how this works.In my view the work down at PV1 should not be equal to half the initial work done because the question states the work done expanding from V1 to V2. Also using the second method wouldn't that still be by definition T2 because although we have algebraically modified the
equation we still have T2 in the formula.

Many thanks

 

[Doc Al]

The method I thought to use is as follows.

PV=nRT we know work = 9.4KJ

The work done equals PΔV, not PV. (When pressure is constant, of course.)

 

[Doc Al]

Also using the second method wouldn't that still be by definition T2 because although we have algebraically modified the
equation we still have T2 in the formula.

There should not be a T2 in the formula. The second method uses PV₁ = nRT₁.

 

[ramzerimar]

Actually, your class mate is right. You said that PV=nRT = 9,4 KJ, and that's not right. Remember that the general definition of work in thermodynamics is the integral of PdV. This is a isobaric process, so the expression turns to: P(Vf-Vi), or PΔV. It's the variation of volume!
Also, you used 4,8 instead of 4,7 for 2PV1= 9.4 KJ (divide by 2). I found 706 K for the initial temperature.