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Question Number 13609 by tawa tawa last updated on 21/May/17

Answered by ajfour last updated on 21/May/17

W=∫_( V_1 ) ^(  V_2 ) PdV  when pressure be constant,  W=P(V_2 −V_1 )       =(10^5 Pa)(12m^3 −10m^3 )       =2×10^5 J  = 200kJ .

$$\mathrm{W}=\int_{\:\mathrm{V}_{\mathrm{1}} } ^{\:\:\mathrm{V}_{\mathrm{2}} } \mathrm{PdV} \\ $$$$\mathrm{when}\:\mathrm{pressure}\:\mathrm{be}\:\mathrm{constant}, \\ $$$$\mathrm{W}=\mathrm{P}\left(\mathrm{V}_{\mathrm{2}} −\mathrm{V}_{\mathrm{1}} \right) \\ $$$$\:\:\:\:\:=\left(\mathrm{10}^{\mathrm{5}} \mathrm{Pa}\right)\left(\mathrm{12m}^{\mathrm{3}} −\mathrm{10m}^{\mathrm{3}} \right) \\ $$$$\:\:\:\:\:=\mathrm{2}×\mathrm{10}^{\mathrm{5}} \mathrm{J}\:\:=\:\mathrm{200kJ}\:. \\ $$

Commented by tawa tawa last updated on 21/May/17

God bless you sir.

$$\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir}. \\ $$

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