Question

In a compressor a mixture of 0.3 kg of carbon dioxide and 0.2 kg of nitrogen is compressed from

$P_1=1bar,T_1=300Kto{P}_2=3bar$ in a polytropic process with n = 1.25.

What is the work of compression in this process?

• A. 18.14 kJ
• B. 27.19 kJ
• C. 34.09 kJ
• D. 39.31 kJ

Answer 1

The correct option is C 34.09 kJ

For an ideal gas, the temperature and pressures at the end states of a polytropic process are related as

$T_2=T_1{\left(\frac{P_2}{P_1}\right)}^{n-1/x}=300\times {\left(\frac{3}{1}\right)}^{0.2}=373.72K$

Work of compression is given as

$=\frac{P_1{V}_1-P_2{V}_2}{n-1}=\frac{m\left(R/m\right)(T_1-T_2)}{n-1}$

$m=m_{C{O}_2}+m_{N_2}=0.3+0.2=0.5kg$

$n_{C{O}_2}=\frac{0.3}{44}=0.0068k-mol$

$n_{N_2}=\frac{0.2}{28}=0.0071k-mol$

$\sum n=n_{C{O}_2}+n_{N_2}=0.0068+0.0071=0.0139k-mol$

Now molecular weight of the mixture

$=\frac{m}{\sum n}=\frac{0.5}{0.0139}=35.97$

$\therefore W_{comp}=\frac{0.5\times \left(\frac{8.317}{35.97}\right(300-373.72\left)\right)}{1.25-1}=-34.09kJ$

-ve sign indicates work is done on the mixture.