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A piston-cylinder device contains 1.329 kg of nitrogen gas at 120 kPa and 27 degree C. The gas is now compressed slowly in a polytropic process during which PV^1.49 = constant. The process ends when the volume is reduced by one-half. Determine the entropy change of nitrogen during this process.

User Nijah
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1 Answer

6 votes

Answer:-0.4199 J/k

Step-by-step explanation:

Given data

mass of nitrogen(m)=1.329 Kg

Initial pressure
\left ( P_1\right )=120KPa

Initial temperature
\left ( T_1\right )=27\degree \approx 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
PV^(1.49)=constant


P_(1)V^(1.49)=
P_2\left ((V)/(2)\right )


P_2=337.066KPa


V_2=
0.493* 10^(-3) m^(3)

and entropy is given by


\Delta s=
C_v \ln \left ((P_2)/(P_1)\right )+
C_p \ln \left ((V_2)/(V_1)\right )

Where,
C_v=
(R)/(\gamma-1)=0.6059


C_p=
(\gamma R)/(\gamma -1)=0.9027

Substituting values we get


\Delta s=
0.6059*\ln \left ((337.066)/(120)\right )+
0.9027 \ln \left ((1)/(2)\right )


\Delta s=-0.4199 J/k

User Daniellee
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