Question

1kg of fluid is expanded reversibly according to the law pv=0.5 where p is in bar and v is in m^3/kg. if the final volume is four times the initial volume.Calculate the work done by the fluid and sketch the process on a pv-diagram

Answer 1

$$W=\displaystyle\int_{V1}^{V2} pdV$$

$$pv=0.5$$

$$p=\frac{0.5}{v}$$

$$1 bar=10^5 Pa$$

$$W=\displaystyle\int_{V1}^{4V1} \frac{0.5\cdot10^5}{V}dV=$$

$$=0.5\cdot10^5\cdot\ln\frac{4}{1}=69000 J$$