Answer to Question #143823 in Mechanical Engineering for Shayan Bhattacharjee

Question

The properties of a closed system change following the relation between pressure and volume as pV= 3.0, where P is in bar V is in m^3. Calculate the work done when the pressure increases from 1.5 bar to 7.5 bar.

Answer 1

The relation between pressure(P) and volume (V) of a closed system is given as

$P\times V=3=constant$

$P=\frac{3}{V}$ ---------------------------------------(1)

$P_1\times V_1=P_2 \times V_2$

when P=1.5 bar=1.5$\times 10^5 Pa$ => V=$\frac{3}{1.5\times 10^5}=2\times 10^{-5}$ $m^3$

when P=7.5 bar=1.5$\times 10^5 Pa$ => V=$\frac{3}{7.5\times 10^5}=0.4\times 10^{-5} m^3$ $m^3$

The work done during the process is given by

$W=\int_{V_1 }^{V_2}PdV=\int_{2\times 10^{-5} }^{0.5\times 10^{-5}}\frac{3}{V}]dV=3\times ln(\frac{0.4}{2})=-3\times 1.6 J=-4.828J$

Negative sign indicates work done on the system.