Question

Refrigerant 22 is at an initial temperature of 60° F and an initial pressure of 80 lb/in^2. The Refrigerant 22 undergoes an isobaric process to a final state of 100° F. Determine the change in specific volume of the Refrigerant 22.

Answer 1

Change in specific volume will be
`0.076ft^3/lb`

Step-by-step explanation:

We have given
`R_(22)` is at 60°F and an initial pressure of
`80lb/in^2`

Now according to
`R_(22)` property at 60°F and at a pressure of
`80lb/in^2` specific volume is 0.7291
`ft^3/lb`

As it is an isobaric process so pressure will be constant that is
`P_1=P_2`

So at 100°F and at 80
`80 lb/in^3` specific volume will be 0.8051
`ft^3/lb`

So change in volume = 0.8051 -0.7291 = 0.076
`ft^3/lb`