Thermodynamics for Mechanical Engineering Problem

[Mi77]

Homework Statement

A closed, rigid tank contains a two-phase liquid-vapor mixture of Refrigerant 22 initially at -20 C with a quality of 50.36%. Energy transfer by heat into the tank occurs until the refrigerant is at a final pressure of 6 bar. Determine the final temperature. If the final state is in the superheated vapor region, at what temperature does the tank contain only saturated vapor?

Homework Equations

v = vf + x(vg-vf)

The Attempt at a Solution

Ok so there are two states in this problem. since I have T1 & x=.5036
I looked at the reference tables in the back of my book and used vf, vg, and x to compute v ( v being specific volume). my computed v :: vf<v<vg which makes since because the ref. 22 is in the two-phase liq-vap region.
Since it is being heated , v and T are increasing correct?
I don't know what to do from here.
since P2 is given, I just assumed at P2 it will be in super-heated vapor state and used linear interpolation to find T using v i found from the initial state and the surrounding T's and v's from the table. but I'm pretty sure that is wrong because v is supposed to increase. Really don't know what I'm doing.

 

[rude man]

How can v increase when the container is rigid? You're assuming specific volume = 1, a fixed number.

 

[Chestermiller]

since P2 is given, I just assumed at P2 it will be in super-heated vapor state and used linear interpolation to find T using v i found from the initial state and the surrounding T's and v's from the table. but I'm pretty sure that is wrong because v is supposed to increase. Really don't know what I'm doing.

You did it correctly. The combined volume per unit mass doesn't change, because the tank is rigid. So v doesn't change.

So, now what is your game plan for doing the second part?

Chet