Answer:
Work = 51,33 kJ
Explanation:
According to the ideal gas equation, the initial volume is calculated as follows:
[tex]V_{i} =RT_{i}m/P_{i}=(0,2870)kJ/kg K*(25+273,15)K*(0,2)kg/(100)kN/m^{2}=0,1711m^{3}[/tex]
Then the work for an isobaric process can be calculated using the following expression:
[tex]W=P(V_{f}-V_{i} )[/tex]
And considering that,
[tex]V_{f}=4V_{i}[/tex]
The work can be calculated as follows:
[tex]W=P(4V_{i}-V_{i})=3PV_{i} =(3)*(100)kn/m^{2}*(0,1711)m^{3}=51,33kJ[/tex]
Answer:
158.57 kJ
Explanation:
We can assume air is an idela gas and the ideal gas law applies.
[tex]P\cdot{v}=m\cdot{R}\cdot{T}[/tex]
And we know that:
[tex]V_1=4\cdot{V_2}[/tex]
WE can assume the process is isothermal
Therefore the work:
W=mRTln(V₂/V₁)
We can determine the constant C as:
[tex]C=P_1/V_1=P_1/(m\cdot{p})[/tex]
[tex]C=100/(0.2\cdot{1000})=0.5[/tex]
[tex]P_2=0.5\cdot{4}/((0.2\cdot{1000})=0.01[/tex]
P₂ is 0.01 kPa
V₂=0.005
V₁=50
[tex]W=0.2\cdot{0.287}\cdot{298}\cdot{9.21}=158.57[/tex]
The total work is 158.57 kJ
