Question

A rigid, insulated vessel is divided into two compartments connected by a valve. Initially, one compartment, occupying one-third of the total volume, contains air at 500oR, and the other is evacuated. The valve is opened and the air is allowed to fill the entire volume. Assuming the ideal gas model with variable specific heats. Determine: a. the final temperature of the air (in oR) b. the amount of specific entropy produced (in Btu/lbm oR)

Answer 1

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 = $T_i$

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

$u_f$ - $u_i$ = 0

$u_f$ = $u_i$

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;

$T_f = T_i$

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 = $\frac{1}{3}$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 = $c_v$In( $\frac{T_2}{T_1}$ ) + R.In( $\frac{V_2}{V_1}$ )

we substitute

Δs = $c_v$In( $\frac{500}{500}$ ) + 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