Answer:
(a) [tex]W_{isoentropic}=8.125\frac{kJ}{mol}[/tex]
(b) [tex]W_{polytropic}=7.579\frac{kJ}{mol}[/tex]
(c) [tex]W_{isothermal}=5.743\frac{kJ}{mol}[/tex]
Explanation:
Hello,
(a) In this case, since entropy remains unchanged, the constant [tex]k[/tex] should be computed for air as an ideal gas by:
[tex]\frac{R}{Cp_{air}}=1-\frac{1}{k} \\\\\frac{8.314}{29.11} =1-\frac{1}{k}\\[/tex]
[tex]0.2856=1-\frac{1}{k}\\\\k=1.4[/tex]
Next, we compute the final temperature:
[tex]T_2=T_1(\frac{p_2}{p_1} )^{1-1/k}=300K(\frac{1000kPa}{100kPa} )^{1-1/1.4}=579.21K[/tex]
Thus, the work is computed by:
[tex]W_{isoentropic}=\frac{kR(T_2-T_1)}{k-1} =\frac{1.4*8.314\frac{J}{mol*K}(579.21K-300K)}{1.4-1}\\\\W_{isoentropic}=8.125\frac{kJ}{mol}[/tex]
(b) In this case, since [tex]n[/tex] is given, we compute the final temperature as well:
[tex]T_2=T_1(\frac{p_2}{p_1} )^{1-1/n}=300K(\frac{1000kPa}{100kPa} )^{1-1/1.3}=510.38K[/tex]
And the isentropic work:
[tex]W_{polytropic}=\frac{nR(T_2-T_1)}{n-1} =\frac{1.3*8.314\frac{J}{mol*K}(510.38-300K)}{1.3-1}\\\\W_{polytropic}=7.579\frac{kJ}{mol}[/tex]
(c) Finally, for isothermal, final temperature is not required as it could be computed as:
[tex]W_{isothermal}=RTln(\frac{p_2}{p_1} )=8.314\frac{J}{mol*K}*300K*ln(\frac{1000kPa}{100kPa} ) \\\\W_{isothermal}=5.743\frac{kJ}{mol}[/tex]
Regards.
