Answer:
Given R=0.596psia.ft3/lbm.R,Tcr=1164.7R,Pcr=3200psia, Iniatial temperature=400°F=859.67R
But
P1V1=RT1 and P2V2=RT2 note before: P1=P2 and V1=2V2
Hence T1=2T=2*759.67=1719.34R
Using saturated steam table at 400°F Psat=247.67psia,the reduced temperature and pressure is Tr=0.738,Pr=0.0773 the compressibility factor z1=0.93
Iniatial specific volume v1=1.925ft^3/lbm
Also v1=2v2=2*1.925=3.8516ft^3/lbm
at the final state pr2=Pr1=0.0773
Pseudo reduced specific volume vr2 =17.76
Using the compressibility factor table for pr2=0.0773 and vr2=17.76
The final temperature is 1631.60°R or 1171.93°F