Answer to Question #156709 in Molecular Physics | Thermodynamics for revanth

Question

In an engine cylinder a gas has a volumetric analysis of 13%CO2, 12.5%O2 and 74.5% N2. The temperature at the beginning of expansion is 9500C and gas mixture expands reversibly through a volume ratio of 8:1. According to the law pV1.2=constant. Find per kg of gas, the work done and the heat flow. Take Cp for CO2=1.235kJ/kg K and O2=1.088kJ/kg K and N2 is 1.172kJ/kg K.

Answer 1

Percentage of mass (m) =

% of CO2=0.18

% of O2 = 0.13

% of N2 = 0.68

Specific heat at constant pressure $=0.18\times C_p1+0.13\times C_p2+0.68\times C_p3$

$C_p=1.17KJ/Kg K$

Gas constant -

$\Rightarrow R=8.314(\frac{0.187}{0.44}+\frac{0.131}{0.44}+\frac{0.682}{0.44})$

$\Rightarrow R=0.2718 KJ/kg/K$

We know that,

$C_p-C_v=R$

$C_v=1.1725-0.271$

$C_v=0.9$

$i) W=\frac{R(T_1-T_2)}{n-1}$

$\frac{T_2}{T_1}=(\frac{v_1}{v_2})^{n-1}=(\frac{1}{8})^{1.2-1} =0.659$

W=566.8KJ/kg

$Q=\Delta U+W=U_2-U_1+W$

$=0.9\times 10^3(806-1223)+566.8\times 10^3$

$Q=191.41kJ/kg$

$ii)$ $S=R\log(\frac{v_2}{v_1}) =0.2718\times 10^3 \log 8$

$S=0.565KJ/kg K$

$S=C_v \log(\frac{T_1}{T_2})$

$=0.9\times 10^3\log(\frac{122.3}{805.9})$

$=0.376$

Change in entropy $(S_2-S_1)=0.19kJ/kg K$