Answer:
a) the final temperature of the air is 500° R
b) the amount of specific entropy produced is 0.0758 Btu/lb-°R
Explanation:
Given the data in the question;
Air at 500° R =
Using first law of thermodynamic;
δQ = dU + W
now, since the vessel is insulated, the transfer is zero, work done also is zero since there is also no external work done.
δQ = dU + W
0 = dU + 0
dU = 0
- = 0
=
hence, change in internal energy is 0
Now, since the ideal internal energy is a function of temperature, the temperature will also remain the same;
F = 500° R
Therefore, the final temperature of the air is 500° R
b)
given that; initial volume is one-third of the total volume
V₁ = V₂
3V₁ = V₂
3 = V₂/V₁
Now, we take the value of gas constant R from air property table; gas constant R = 0.069 Btu/lb-R
so we calculate the entropy change;
Δs = In( ) + R.In( )
we substitute
Δs = In( ) + 0.069 × In( 3 )
Δs = 0 + [0.069 × In( 3 )]
Δs = 0 + [0.069 × 1.0986]
Δs = 0.0758 Btu/lb-°R
Therefore, the amount of specific entropy produced is 0.0758 Btu/lb-°R