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Alex Ar [27]
3 years ago
14

A piston-cylinder device contains Helium gas initially at 150 kPa, 20 o C, and 0.5m 3 . The helium is now compressed in a polytr

opic process () to 400 kPa and 140 o C. Determine the work and heat transfer done during this process.
Physics
1 answer:
Molodets [167]3 years ago
3 0

Answer:

Explanation:

Given

P_1=150 kPa

T_1=20^{\circ}C

V_1=0.5 m^3

T_2=140^{\circ}C

P_2=400 kPa

R for Helium R=2.076

c_v=3.115 kJ/kg-K

mass of gas m=\frac{P_1V_1}{RT_1}

m=\frac{150\times 0.5}{2.076\times 293}

m=0.123 kg

Similarly V_2 can be found

\frac{P_1V_1}{T_1}=\frac{P_2V_2}{T_2}

V_2=0.264 m^3

Work done W=\int_{V_1}^{V_2}PdV

W=\frac{P_2V_2-P_1V_1}{n-1}

W=\frac{mR(T_2_T_1)}{n-1}

Since it is a polytropic Process

therefore PV^n=c

P_1V_1^n=P_2V_2^n

(\frac{V_1}{V_2})^n=\frac{P_2}{P_1}

(\frac{0.5}{0.264})^n=\frac{400}{150}

n=\frac{\ln 2.66}{\ln 1.893}

n=1.533

W=\frac{0.123\times 2.076(140-20)}{1.533-1}

W=57.48 kJ    

From Energy balance

E_{in}-E_{out}=\Delta E_{system}

Neglecting kinetic and Potential Energy change

Q_{in}+W_{in}=change\ in\ Internal\ Energy

Change in Internal Energy \Delta U=u_2-u_1

\Delta U=mc_v(T_2-T_1)

\Delta U=0.123\times 3.115(140-20)

\Delta U=45.977 kJ

Q_{in}+57.48=45.977

Q_{in}=-11.50 kJ  

i.e. Heat is being removed

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