Mass of carbon dioxide, m = 0.04 kg

 

Molecular weight, M = 44

 

Initial pressure, \(p_{1} = 1\) bar \(= 1 × 10^{5} N/m^{2}\)

 

Initial temperature, \(T_{1} = 20 + 273 = 293 \) K

 

Final pressure, \(p_{2} = 9 \) bar

 

Final volume, \(V_{2} = 0.003 m^{3}\)

 

\(c_{p} \) for carbon dioxide = 0.88 kJ/kg K

 

Change of entropy :

 

Characteristics gas constant,

 

\(R=\frac{R_{0}}{M}=\frac{8314}{44}=189 \) Nm/kg K

 

To find \(T_{2}\), using the relation,

 

\(p_{2}V_{2} = mRT_{2}\)

 

\(\therefore\) \( T_{2}=\frac{p_{2}V_{2}}{mR}=\frac{9\times 10^{5}\times 0.003}{0.04\times 189}=357 \) K

 

Now \(s_{A} – s_{2} = R \log_{e}\frac{p_{2}}{p_{1}}=\frac{189}{10^{3}}\log_{e}\left ( \frac{9}{1} \right )\)

 

= 0.4153 kJ/kg K

 

Also at constant pressure from 1 to A

 

\(s_{A}=s_{1}=c_{p}\log_{e}\frac{T_{2}}{T_{1}}=0.88\log_{e}\left ( \frac{357}{293} \right )\)

 

= 0.1738 kJ/kg K

 

Then \((s_{1} – s_{2}) = (s_{A} – s_{2}) – (s_{A} – s_{1})\)

 

= 0.4153 – 0.1738 = 0.2415 kJ/kg K

 

Hence for 0.04 kg of carbon dioxide decrease in entropy,

 

\(S_{1} – S_{2}= m(s_{1} – s_{2}) = 0.04 × 0.2415\)

 

= 0.00966 kJ/K.

 

Note. In short, the change of entropy can be found by using the following relation :

 

\(\left ( s_{2}-s_{1} \right )=c_{p}\log_{e}\frac{T_{2}}{T_{1}}-R\log_{e}\frac{p_{2}}{p_{1}}=0.88\log_{e}\left ( \frac{357}{293} \right )-\frac{189}{10^{3}}\log_{e}\left ( \frac{9}{1} \right )\)

 

= 0.1738 – 0.4153 = – 0.2415 kJ/kg K

 

\(\therefore\) \( S_{2} – S_{1} = m(s_{2} – s_{1}) = 0.04 × (– 0.2415)\)

 

= – 0.00966 kJ/K

 

 

(–ve sign means decrease in entropy)

 

or \(S_{1} – S_{2} = 0.00966\) kJ/K.