Answer to Question #295917 in Molecular Physics | Thermodynamics for dandan

Question

f 10 kg/min of air are compressed isothermally from P1=96kPa. Find the work change entropy and the heat for

(A) non flow process with v1=15m/s and v2=60m/s

Answer 1

Solution;

Given;

$\dot{m}=10kg/min=0.16667kg/s$

$P_1=96kPa$

$V_1=15m/s$

$V_2=60m/s$

$V_2=620kPa$

$v_1=7.65m^3/min=0.1275m^3/s$s

Work;

$W=nRT_1ln(\frac{P_1}{P_2})$

$n=0.3452$

$R=8.314J/molK$

$T_1=\frac{PV}{n}=\frac{96×0.1275}{0.3452}=35.45K$

Hence;

$W=0.3452×8.314×35.45ln(\frac{96}{620})=-190.33kJ$

Heat;

$Q=U+W$

But U=0

$Q=W=-190kJ$

Entropy;

$S=\frac{Q}{T_1}=\frac{-190.33}{35.45}=-5.36kJ$