Answer to Question #102719 in Molecular Physics | Thermodynamics for MAHITI JULIUS

Question

0.3kg of air at pressure of 350kN/m2 and a temperature of 35 oC receives heat energy at constant volume until its pressure become 700kN/m2. it then receives heat energy at constant pressure until its volume become 0.2289m3. determine the change of entropy during each process. take Cp = 1.006kJ/kgK and Cv = 0.717kJ/kgK

Answer 1

$\delta Q=\Delta U+pdV=\frac{m}{M}C_VdT+pdV$

If $V=const$

$\delta Q=\Delta U+pdV=\frac{m}{M}C_VdT$

$\Delta S_1 =\int\frac{\delta Q}{T}=\int \frac{(m/M)C_VdT }{T}=\frac{m}{M}C_V\ln \frac{T_2}{T_1}=\frac{m}{M}C_V\ln \frac{p_2}{p_1}=\frac{0.3}{28.96\cdot 10^{-3}}\cdot 717\cdot\ln \frac{700\cdot 10^{-3}}{350\cdot 10^{-3}}=5148 J/K$

if $p=const$

$\delta Q=\Delta U+pdV=\frac{m}{M}C_VdT+pdV=\frac{m}{M}C_pdT$

$\Delta S_2 =\int\frac{\delta Q}{T}=\int \frac{(m/M)C_pdT}{T} = \frac{m}{M}C_p\ln \frac{T_2}{T_1}=\frac{m}{M}C_p\ln \frac{V_2}{V_1}$

$V_1=\frac{(m/M)RT_1}{p_1}=\frac{(0.3/0.02896)\cdot 8.31\cdot 308}{350\cdot 10^{3}}=0.07575m^3$

$\Delta S_2 =\frac{m}{M}C_p\ln \frac{V_2}{V_1}=\frac{0.3}{0.02896} \cdot 1006\cdot \ln \frac{0.2289}{0.07575}=11524 J/K$

Answer. $\Delta S_1=5148 J/K; \Delta S_2=11524 J/K$