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Question 11.140: A 0.2 m^3 insulated, rigid vessel is divided into two equal ...

A 0.2 m ^{3} insulated, rigid vessel is divided into two equal parts A and B by an insulated partition, as shown in Fig. P11.140. The partition will support a pressure difference of 400 kPa before breaking. Side A contains methane and side B contains carbon dioxide. Both sides are initially at 1 MPa, 30°C. A valve on side B is opened, and carbon dioxide flows out. The carbon dioxide that remains in B is assumed to undergo a reversible adiabatic expansion while there is flow out. Eventually the partition breaks, and the valve is closed. Calculate the net entropy change for the process that begins when the valve is closed.

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\begin{aligned}&\Delta P _{ MAX }=400   kPa , \quad P _{ A1 }= P _{ B 1}=1   MPa \\& V _{ A1}= V _{ B 1}=0.1   m ^{3} \\& T _{ A 1}= T _{ B 1}=30^{\circ} C =303.2   K\end{aligned}

 

CO _{2}  inside B:  s _{ B 2}= s _{ B 1} \text { to } P _{ B 2}=600   kPa \left( P _{ A 2}=1000   kPa \right)

For  CO _{2}  ,        k =1.289 \Rightarrow T _{ B 2}=303.2\left(\frac{600}{1000}\right)^{\frac{0.289}{1.289}}=270.4   K

 

\begin{aligned}& n _{ B 2}= P _{ B 2} V _{ B 2} / \overline{ R } T _{ B 2}=600 \times 0.1 / 8.3145 \times 270.4=0.026688 \\& n _{ A 2}= n _{ A1 }=1000 \times 0.1 / 8.3145 \times 303.2=0.039668   kmol\end{aligned}

 

The process 2 to 3 is adiabatic but irreversible with no work.

\begin{gathered}Q _{23}=0= n _{3} \overline{ u }_{3}-\Sigma_{ i } n _{ i 2} \overline{ u }_{ i 2}+0= n _{ A 2} \overline{ C }_{ vo  A }\left( T _{3}- T _{ A 2}\right)+ n _{ B 2} \overline{ C }_{ vo  B }\left( T _{3}- T _{ B 2}\right)=0 \\0.039668 \times 16.04 \times 1.736\left( T _{3}-303.2\right)+0.026688 \times 44.01 \times 0.653\left( T _{3}-270.4\right)=0\end{gathered}

Solve      T _{3}=289.8   K

Get total and partial pressures for the entropy change

\begin{aligned}& P _{3}= n \overline{ R } T / V =0.066356 \times 8.3145 \times 289.8 / 0.2=799.4   kPa \\& P _{ A 3}=0.5978 \times 799.4=477.9   kPa , \quad P _{ B 3}= P _{3}- P _{ A 3}=321.5   kPa\end{aligned}

 

\begin{aligned}&\overline{ s }_{ A 3}-\overline{ s }_{ A 2}=16.04 \times 2.254 \ln \left(\frac{289.8}{303.2}\right)-8.3145 \ln \frac{477.9}{1000}=4.505   kJ / kmol  K \\&\overline{ s }_{ B 3}-\overline{ s }_{ B 2}=44.01 \times 0.842 \ln \left(\frac{289.8}{270.4}\right)-8.3145 \ln \frac{321.5}{600}=7.7546   kJ / kmol  K\end{aligned}

 

\begin{aligned}\Delta S _{ NET } &= n _{ A 2}\left(\overline{ s }_{ A 3}-\overline{ s }_{ A 2}\right)+ n _{ B 2}\left(\overline{ s }_{ B 3}-\overline{ s }_{ B 2}\right) \\&=0.039668 \times 4.505+0.026688 \times 7.7546=+ 0 . 3 8 5 7   k J / K\end{aligned}

 

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