Question

A piston cylinder containing 0.2 kg of air undergoes a process where pressure and volume are related by the expression P=CV where C is a constant value. If the initial pressure is 0.1 MPa and the initial temperature is 25 °C, determine the work done in units of kJ for the final volume to be four times the initial volume.

Answer 1

158.57 kJ

Step-by-step explanation:

We can assume air is an idela gas and the ideal gas law applies.

`P\cdot{v}=m\cdot{R}\cdot{T}`

And we know that:

`V_1=4\cdot{V_2}`

WE can assume the process is isothermal

Therefore the work:

W=mRTln(V₂/V₁)

We can determine the constant C as:

`C=P_1/V_1=P_1/(m\cdot{p})`

`C=100/(0.2\cdot{1000})=0.5`

`P_2=0.5\cdot{4}/((0.2\cdot{1000})=0.01`

P₂ is 0.01 kPa

V₂=0.005

V₁=50

`W=0.2\cdot{0.287}\cdot{298}\cdot{9.21}=158.57`

The total work is 158.57 kJ

Answer 2

Work = 51,33 kJ

Step-by-step explanation:

According to the ideal gas equation, the initial volume is calculated as follows:

`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)`

Then the work for an isobaric process can be calculated using the following expression:

`W=P(V_(f)-V_(i) )`

And considering that,

`V_(f)=4V_(i)`

The work can be calculated as follows:

`W=P(4V_(i)-V_(i))=3PV_(i) =(3)*(100)kn/m^(2)*(0,1711)m^(3)=51,33kJ`