Answer:
Part a: <em>The total amount of energy transfer by the work done is 54.81 kJ.</em>
Part b: <em>The total amount of energy transfer by the heat is 54.81 kJ</em>
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
is given as
M=44 kg/kmol
the gas constant is given as
![R=\frac{\bar{R}}{M}\\R=\frac{8.314}{44}\\R=0.189 kJ/kg.K](https://tex.z-dn.net/?f=R%3D%5Cfrac%7B%5Cbar%7BR%7D%7D%7BM%7D%5C%5CR%3D%5Cfrac%7B8.314%7D%7B44%7D%5C%5CR%3D0.189%20kJ%2Fkg.K)
Volume of the tank is given as
![V=\frac{mRT}{P_1}\\V=\frac{3 \times 0.189 \times 290}{300 }\\V=0.5481 m^3](https://tex.z-dn.net/?f=V%3D%5Cfrac%7BmRT%7D%7BP_1%7D%5C%5CV%3D%5Cfrac%7B3%20%5Ctimes%200.189%20%5Ctimes%20290%7D%7B300%20%7D%5C%5CV%3D0.5481%20m%5E3)
Final mass is given as
![m_2=\frac{P_2V}{RT}\\m_2=\frac{200\times 0.5481}{0.189\times 290}\\m_2=2 kg](https://tex.z-dn.net/?f=m_2%3D%5Cfrac%7BP_2V%7D%7BRT%7D%5C%5Cm_2%3D%5Cfrac%7B200%5Ctimes%200.5481%7D%7B0.189%5Ctimes%20290%7D%5C%5Cm_2%3D2%20kg)
Mass of the CO2 moved to the cylinder
![m=m_1-m_3\\m=3-2=1 kg](https://tex.z-dn.net/?f=m%3Dm_1-m_3%5C%5Cm%3D3-2%3D1%20kg)
The initial mass in the cylinder is given as
![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](https://tex.z-dn.net/?f=m_%7B%28cyl%29_1%7D%3D%5Cfrac%7BP_%7B%28cyl%29_1%7DV_1%7D%7BRT%7D%5C%5Cm_%7B%28cyl%29_1%7D%3D%5Cfrac%7B200%5Ctimes%200.5%7D%7B0.189%20%5Ctimes%20290%7D%5C%5Cm_%7B%28cyl%29_1%7D%3D1.82%20kg)
The mass after the process is
![m_{(cyl)_2}=m_{(cyl)_1}+m\\m_{(cyl)_2}=1.82+1\\m_{(cyl)_2}=2.82\\](https://tex.z-dn.net/?f=m_%7B%28cyl%29_2%7D%3Dm_%7B%28cyl%29_1%7D%2Bm%5C%5Cm_%7B%28cyl%29_2%7D%3D1.82%2B1%5C%5Cm_%7B%28cyl%29_2%7D%3D2.82%5C%5C)
Now the volume 2 of the cylinder is given as
![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](https://tex.z-dn.net/?f=V_%7B%28cyl%29_2%7D%3D%5Cfrac%7Bm_%7B%28cyl%29_2%7DRT%7D%7BP_2%7D%5C%5Cm_%7B%28cyl%29_2%7D%3D%5Cfrac%7B2.82%5Ctimes%200.189%5Ctimes%20290%7D%7B200%7D%5C%5Cm_%7B%28cyl%29_1%7D%3D0.774%20m%5E3)
Part a:
So the Work done is given as
![W=P(V_2-V_1)\\W=200(0.774-0.5)\\W=54.81 kJ](https://tex.z-dn.net/?f=W%3DP%28V_2-V_1%29%5C%5CW%3D200%280.774-0.5%29%5C%5CW%3D54.81%20kJ)
<em>The total amount of energy transfer by the work done is 54.81 kJ.</em>
Part b:
The total energy transfer by heat is given as
![Q=\Delta U+W\\Q=0+W\\Q=54.81 kJ](https://tex.z-dn.net/?f=Q%3D%5CDelta%20U%2BW%5C%5CQ%3D0%2BW%5C%5CQ%3D54.81%20kJ)
As the temperature is constant thus change in internal energy is 0.
<em>The total amount of energy transfer by the heat is 54.81 kJ</em>