Answer:
[tex]W=46141.72~J[/tex]
Explanation:
Given:
initial pressure, [tex]P_1=14~bar=1.4\times10^6~Pa[/tex]
initial volume, [tex]V_1=0.03~m^3[/tex]
After isothermal expansion:
final volume, [tex]V_2=3V_1=0.09~m^3[/tex]
a)
We have the work done in isothermal process as:
[tex]W=P_1.V_1\ln(\frac{V_2}{V_1} )[/tex]
[tex]W=1.4\times10^6\times 0.03\ln(\frac{0.09}{0.03} )[/tex]
[tex]W=46141.72~J[/tex] is the work by the gas in moving the external force
b)
When the force on piston is suddenly reduced to half the initial value then the process becomes near to adiabatic.
[tex]P_1V_1^{\gamma}=P_2V_2^{\gamma}[/tex]
[tex]14\times 0.03^{1.4}=P_2\times 0.09^{1.4}[/tex]
[tex]P_2=3~bar[/tex]
Now, using the formula for work done in adiabatic process:
[tex]W=\frac{P_1V_1-P_2V_2}{\gamma-1}[/tex]
[tex]W=\frac{1.4\times10^6\times 0.03-3\times10^5\times 0.09}{1.4-1}[/tex]
[tex]W=3750~J[/tex]
