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

Answer:

158.57 kJ

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

Answer:

Work = 51,33 kJ

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:

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