Question Number 13609 by tawa tawa last updated on 21/May/17
Answered by ajfour last updated on 21/May/17
$$\mathrm{W}=\int_{\:\mathrm{V}_{\mathrm{1}} } ^{\:\:\mathrm{V}_{\mathrm{2}} } \mathrm{PdV} \\ $$$$\mathrm{when}\:\mathrm{pressure}\:\mathrm{be}\:\mathrm{constant}, \\ $$$$\mathrm{W}=\mathrm{P}\left(\mathrm{V}_{\mathrm{2}} −\mathrm{V}_{\mathrm{1}} \right) \\ $$$$\:\:\:\:\:=\left(\mathrm{10}^{\mathrm{5}} \mathrm{Pa}\right)\left(\mathrm{12m}^{\mathrm{3}} −\mathrm{10m}^{\mathrm{3}} \right) \\ $$$$\:\:\:\:\:=\mathrm{2}×\mathrm{10}^{\mathrm{5}} \mathrm{J}\:\:=\:\mathrm{200kJ}\:. \\ $$
Commented by tawa tawa last updated on 21/May/17
$$\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir}. \\ $$