Answer to Question #256848 in Physics for freebandz

Question

1) Derive the expression of workdone by the position on the cylinder in terms of the pressure and the volume of fluid in the cylinder

2) A unit of a fluid at a pressure of 3 bar and with a specific volume of 0.18m³/kg contained in a cylinder behind a piston that expands reversibly to a pressure of 0.6 bar the equation of pressure is taken as below p= c/v² where c is a constant. Calculate the workdone during the process.

Answer 1

1) The work is force times change in position caused by the action of the force, right?

$W=F\Delta h.\\ F=pA,$

where A - area:

$W=pA\Delta h=p\Delta V.$

2) The work is

$W=c\int^{v_2}_{v_1}\frac{dv}{v^2}=c\bigg(\frac1{v_1}-\frac1{v_2}\bigg).$

Since the product of pressure and specific volume is constant:

$c=p_1v_1^2=p_2v_2^2,\\c=3·0.18^2=0.0972\text{ bar}·(\text{ m}^3/\text{kg})^2.\\\space\\ v_2=\sqrt{c/p^2}=\sqrt{0.0972/0.6}=0.402\text{ m}^3/\text{kg}.$

Therefore:

$W=0.0972·10^5\bigg(\frac1{0.18}-\frac1{0.402}\bigg)=29821\text{ J}.$