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Solve Problem 35.4 assuming that the process is adiabatic. |
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Answer» `(dgamma)/((d//x)^(gamma))=(1+(x)/(d))^(gamma)~~1-(gammax)/(d),(d^(gamma))/((d-x)^(gamma))=(1-(x)/(d))^(gamma)~~1+(gammax)/(d)` For the force we obtain `F=(p_(1)-p_(2))S=[(PD^(gamma))/((d-x)^(gamma))-(pd^(gamma))/((d+x)^(gamma))]S~~(2gammapSx)/(d)=(2gammapVx)/(d^(2))` i.e. `F_(ad)=gammaF_("isot")`. |
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