Question

A rigid tank initially contains 3kg of carbon dioxide (CO2) at a pressure of 3bar.The tank is connected by a valve to a frictionless piston-cylinder assembly located vertically above, initially containing 0.5 m^3 of CO2. The piston area is 0.1 m^2. Initially the pressure of the CO2 in the piston-cylinder assembly is 2 bar. The ambient pressure and temperature are 1 bar and 290 K. Although the valveis closed, a small leak allows CO2 to flow slowly into the cylinder from the tank. Owing to heat transfer, the temperature of the CO2 throughout the tank and the piston-cylinder assembly stays constant at 290K. You can assume ideal gas behavior for CO2.
Determine the following:

a. The total amount of energy transfer by work (kJ)
b. The total amount of energy transfer by heat (kJ)

Answer 1

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 $CO_2$ is given as

M=44 kg/kmol

the gas constant is given as

$R=\frac{\bar{R}}{M}\\R=\frac{8.314}{44}\\R=0.189 kJ/kg.K$

Volume of the tank is given as

$V=\frac{mRT}{P_1}\\V=\frac{3 \times 0.189 \times 290}{300 }\\V=0.5481 m^3$

Final mass is given as

$m_2=\frac{P_2V}{RT}\\m_2=\frac{200\times 0.5481}{0.189\times 290}\\m_2=2 kg$

Mass of the CO2 moved to the cylinder

$m=m_1-m_3\\m=3-2=1 kg$

The initial mass in the cylinder is given as

$m_{(cyl)_1}=\frac{P_{(cyl)_1}V_1}{RT}\\m_{(cyl)_1}=\frac{200\times 0.5}{0.189 \times 290}\\m_{(cyl)_1}=1.82 kg$

The mass after the process is

$m_{(cyl)_2}=m_{(cyl)_1}+m\\m_{(cyl)_2}=1.82+1\\m_{(cyl)_2}=2.82$

Now the volume 2 of the cylinder is given as

$V_{(cyl)_2}=\frac{m_{(cyl)_2}RT}{P_2}\\m_{(cyl)_2}=\frac{2.82\times 0.189\times 290}{200}\\m_{(cyl)_1}=0.774 m^3$

Part a:

So the Work done is given as

$W=P(V_2-V_1)\\W=200(0.774-0.5)\\W=54.81 kJ$

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

$Q=\Delta U+W\\Q=0+W\\Q=54.81 kJ$

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