A vessel is containing a saturated mixture of 1kg liquid water and 100g steam at 100 degrees celcius.

The vessel is connected to another container with the exact same volume, and the mixture expands freely so it fills the doubled volume. The system is thermally isolated.

What is the final temperature of the mixture after expansion?

What i have done: tried to find specific volume, x=mg/m, where mg is 100g(0.1kg) steam and m= mg*m= 0.1(100g steam)*1(1kg water)=0.1

x=0.1/0.1 = 1. Since the volume doubled = 2*1 = 2

v=vf*x(vg-vf) where is x is 2 as stated above, v=0.001043*2(1.6720-0.001043) = 3.342 m^3/kg

Not sure about have to find specific internal energy, since no heat or and work is added. so i just found the table values, vf=419.06 and vg=2506.08 and evap=2087.

and how to solve to find final temperature i dont know.

any help would be appreciated

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