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A sample of ideal gas is expanded to twice its original volume of `1.00m^3` in a quasi-static process for which `p=alphaV^2`, with `alpha=5.00 atm//m^6`, as shown in Fig. How much work is done by the expanding gas? |
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Answer» Correct Answer - A `W=int_(V_i)^(V_f)pdV=int_1^2(alphaV^2)dV` `=int_1^2(5xx1.01xx10^5)V^2dV` Solving we get, `W=1.18xx10^6J` `=1.18MJ` |
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