Guest Question.

0.04 kg of carbon dioxide (molecular weight = 44) is compressed from 1 bar,20°C, until the pressure is 9 bar, and the volume is then 0.003 $$m^{3}$$. Calculate the change of entropy.Take $$c_{p}$$ for carbon dioxide as 0.88 kJ/kg K, and assume carbon dioxide to be a perfect gas.

Solution

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.