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Image for 2. A vessel with a piston cylinder attached to a spring (see Figure 1) contains CO2 with a volume of 500 L, a

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Answer #1

state 1

V_{1}=500L

P_{1}=200kPa

T_{1}=25^{\circ}C=298 K

R=8.314 LkPaK^{-1}mol^{-1}

using ideal gass equation

PV=nRT

n=\frac{PV}{RT}

\Rightarrow n=40.362 mol

mass of CO2 per mole is 44 gram

total mass=n*44 gram

total mass=1775.928 gram

total mass=1.776 Kg

state 2

V_{2}=let x L

P_{2}=400kPa

T_{1}=500^{\circ}C=773 K

R=8.314 LkPaK^{-1}mol^{-1}

using ideal gass equation

PV=nRT

V=\frac{nRT}{P}

V_{2}=648.49L=0.648 m^{3}

state 3

V_{3}=648.49L

P_{3}=600kPa

T_{1}=let x K

R=8.314 LkPaK^{-1}mol^{-1}

using ideal gass equation

PV=nRT

T=\frac{PV}{nR}

T_{3}=1159.5K=886.5^{\circ}C

taking expternal pressure equal to 1 Atm=101.32kPa

boundry work

W=-P_{ext}\Delta V

W=-P_{ext}(V_{3}-V_{1})

W=-15045 kPaL=-15.045kJ

Q_{23}=mC_{p}(T_{3}-T_{2})

Q_{23}=170.33KJ

now change in internal energi from 1-2

\Delta U=mC_{v}(T_{3}-T_{1})

\Delta U=686.68KJ

Q_{total}=\Delta U+W

Q_{total}=\Delta U+W

Q_{total}=(686.68-15.045)KJ

Q_{total}=671.63KJ

Q_{total}=Q_{12}+Q_{23}

Q_{12}=Q_{total}-Q_{23}

Q_{12}=501.30KJ

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