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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

$$\mathrm{W}={\int }_{{\mathrm{V}}_1}^{{\mathrm{V}}_2}\mathrm{PdV}$$$$\mathrm{when}\mathrm{pressure}\mathrm{be}\mathrm{constant},$$$$\mathrm{W}=\mathrm{P}({\mathrm{V}}_2-{\mathrm{V}}_1)$$$$=\left({10}^5\mathrm{Pa}\right)({12\mathrm{m}}^3-{10\mathrm{m}}^3)$$$$=2\times {10}^5\mathrm{J}=200\mathrm{kJ}.$$

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

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

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