Question

A gas enclosed in a cylinder-piston arrangement is heated slowly to change its state from 100 kPa, 0.03 m3 to an initial state of 0.1 m3. Assuming volume of the gas changes inversely with pressure, determine amount of work done during the process.

Answer 1

$W=\int_{1}^{2} P\Delta V$ Since pressure is constant, $W=P\int_{1}^{2} \Delta V=P(v_1-V_2)$.

Therefore, Work-done$= 100kPa(0.1-0.03) =7.0kJ$

Work done =7.0kJ.