Answer:-0.4199 J/k
Step-by-step explanation:
Given data
mass of nitrogen(m)=1.329 Kg
Initial pressure
=120KPa
Initial temperature
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
=constant
=
![P_2\left ((V)/(2)\right )](https://img.qammunity.org/2020/formulas/engineering/college/zsorv0w6i9qcwro8rvoi2dzp0jfyrgfx8c.png)
=337.066KPa
=
![0.493* 10^(-3) m^(3)](https://img.qammunity.org/2020/formulas/engineering/college/rrvw073szsq1zobvsntdj9dnanagkbize4.png)
and entropy is given by
=
+
![C_p \ln \left ((V_2)/(V_1)\right )](https://img.qammunity.org/2020/formulas/engineering/college/heyguu8r3ebge2dmxghz1bqafvgd9rs8dl.png)
Where,
=
=0.6059
=
=0.9027
Substituting values we get
=
+
![0.9027 \ln \left ((1)/(2)\right )](https://img.qammunity.org/2020/formulas/engineering/college/9hpqfp1r7uxrkcgefr9zdb9dml5mer1i8z.png)
=-0.4199 J/k