Reversible Isothermal Expansion Steam

[zwambooo]

Homework Statement

2kgs (total mass) of steam goes through a revesible isothermal expansion at 500 degrees celsius. During the expansion the pressure drops from 300 kpa to 200 kpa.
What is the heat absorbed by the steam during this process?

Homework Equations

U=W and W=nrt ln(v2/v1)

The Attempt at a Solution

Steam cannot be assumed to be an ideal gas atleast to my knowledge. When I google Reversible Isothermal Expansion all i get is that W is given by, nRT*ln(v2/v1). I cannot imagine the W=nRT*ln(v2/v1) being used in this specific problem. This formula would not give a numerical answer anyway, because the problem states no information about volumes.In a isothermal expansion: there is no change in internal energy so it is 0, and the work is done by the system so W is positive Q=+W.. So how do i calculate this W(heat absorbed by the steam) inthis case? is there any magical formula:)

thank you for all input:)!

 

[Chestermiller]

Under these conditions, steam can be approximated as an ideal gas. What does the ideal gas law tell you about the volume ratio of the process is isothermal and you know the pressure ratio?

 

[zwambooo]

Under these conditions, steam can be approximated as an ideal gas. What does the ideal gas law tell you about the volume ratio of the process is isothermal and you know the pressure ratio?

ahhh, so (v2/v1) = (p1/p2)? so W=nRT*ln(p1/p2) in this case