Question

0.05 m^3 of gas at 6.9 bar expands reversibly in a Cylinder behind a piston according to a PV^ prime prime =C until the Volume is 0.08 m Calculate the work done by the gas and Sketch the process 0.05m ^ 3 of law on p-v diagram.

Answer 1

From the above $PV^{2} =C$ where C is constant

1 bar = $10^{5}N/m^{2}$

from the above $p_{1}$ = $6.9 \times10^{5}N/m^{2}$

Work done (W)=

$W=\int_{v_{1}}^{v_{2}}p\partial v$

$W=\int_{v_{1}}^{v_{2}}\frac{C}{v^{2}}\partial v =C\int_{v1}^{v2}\frac{\partial v}{v^{2}}\\ C\left [ \frac{v^{-2+1}}{-2+1} \right ] from\ _{v_{2}}\to v_{1}\\$

$C=\left [ \frac{1}{v_{1}}-\frac{1}{v_{2}} \right ]$

$C= pv^{2}=p_{1}v^{2}_{1}=6.9\times10^{5}N/m^{2}\times0.05m^{2}$

=34500

Substituting the values of $v_{1},v_{2}$ and C in the equation above we get

$W=34500 \left [ \frac{1}{0.05} -\frac{1}{0.08}\right ]N/m$

Work done =258750 N/m