Answer:
Part a: The total amount of energy transfer by the work done is 54.81 kJ.
Part b: The total amount of energy transfer by the heat is 54.81 kJ
Explanation:
Mass of Carbon Dioxide is given as m1=3 kg
Pressure is given as P1=3 bar =300 kPA
Volume is given as V1=0.5 m^3
Pressure in tank 2 is given as P2=2 bar=200 kPa
T=290 K
Now the Molecular weight of [tex]CO_2[/tex] is given as
M=44 kg/kmol
the gas constant is given as
[tex]R=\frac{\bar{R}}{M}\\R=\frac{8.314}{44}\\R=0.189 kJ/kg.K[/tex]
Volume of the tank is given as
[tex]V=\frac{mRT}{P_1}\\V=\frac{3 \times 0.189 \times 290}{300 }\\V=0.5481 m^3[/tex]
Final mass is given as
[tex]m_2=\frac{P_2V}{RT}\\m_2=\frac{200\times 0.5481}{0.189\times 290}\\m_2=2 kg[/tex]
Mass of the CO2 moved to the cylinder
[tex]m=m_1-m_3\\m=3-2=1 kg[/tex]
The initial mass in the cylinder is given as
[tex]m_{(cyl)_1}=\frac{P_{(cyl)_1}V_1}{RT}\\m_{(cyl)_1}=\frac{200\times 0.5}{0.189 \times 290}\\m_{(cyl)_1}=1.82 kg[/tex]
The mass after the process is
[tex]m_{(cyl)_2}=m_{(cyl)_1}+m\\m_{(cyl)_2}=1.82+1\\m_{(cyl)_2}=2.82\\[/tex]
Now the volume 2 of the cylinder is given as
[tex]V_{(cyl)_2}=\frac{m_{(cyl)_2}RT}{P_2}\\m_{(cyl)_2}=\frac{2.82\times 0.189\times 290}{200}\\m_{(cyl)_1}=0.774 m^3[/tex]
Part a:
So the Work done is given as
[tex]W=P(V_2-V_1)\\W=200(0.774-0.5)\\W=54.81 kJ[/tex]
The total amount of energy transfer by the work done is 54.81 kJ.
Part b:
The total energy transfer by heat is given as
[tex]Q=\Delta U+W\\Q=0+W\\Q=54.81 kJ[/tex]
As the temperature is constant thus change in internal energy is 0.
The total amount of energy transfer by the heat is 54.81 kJ
