Answer:-0.4199 J/k
Explanation:
Given data
mass of nitrogen(m)=1.329 Kg
Initial pressure[tex]\left ( P_1\right )[/tex]=120KPa
Initial temperature[tex]\left ( T_1\right )=27\degree \approx[/tex] 300k
Final volume is half of initial
R=particular gas constant
Therefore initial volume of gas is given by
PV=mRT
V=0.986\times 10^{-3}
Using [tex]PV^{1.49}[/tex]=constant
[tex]P_{1}V^{1.49}[/tex]=[tex]P_2\left (\frac{V}{2}\right )[/tex]
[tex]P_2[/tex]=337.066KPa
[tex]V_2[/tex]=[tex]0.493\times 10^{-3} m^{3}[/tex]
and entropy is given by
[tex]\Delta s[/tex]=[tex]C_v \ln \left (\frac{P_2}{P_1}\right )[/tex]+[tex]C_p \ln \left (\frac{V_2}{V_1}\right )[/tex]
Where, [tex]C_v[/tex]=[tex]\frac{R}{\gamma-1}[/tex]=0.6059
[tex]C_p[/tex]=[tex]\frac{\gamma R}{\gamma -1}[/tex]=0.9027
Substituting values we get
[tex]\Delta s[/tex]=[tex]0.6059\times\ln \left (\frac{337.066}{120}\right )[/tex]+[tex]0.9027 \ln \left (\frac{1}{2}\right )[/tex]
[tex]\Delta s[/tex]=-0.4199 J/k
