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(A) Given the differential equation $\frac{dp}{dv} = -ap$. Integrating both sides: $\int_{p_{0}}^{p} \frac{dp}{p} = -a \int_{0}^{v} dv$. This yields $

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$\frac{p_{0}}{aeR}$

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(A) Given the differential equation $\frac{dp}{dv} = -ap$. Integrating both sides: $\int_{p_{0}}^{p} \frac{dp}{p} = -a \int_{0}^{v} dv$. This yields $\ln(\frac{p}{p_{0}}) = -av$,so $p = p_{0}e^{-av}$. For one mole of an ideal gas,$PV = RT$,so $T = \frac{PV}{R} = \frac{p_{0}ve^{-av}}{R}$. To find the maximum temperature,differentiate $T$ with respect to $v$ and set it to zero: $\frac{dT}{dv} = \frac{p_{0}}{R} [e^{-av} + v(-a)e^{-av}] = \frac{p_{0}e^{-av}}{R} (1 - av) = 0$. This gives $v = \frac{1}{a}$. Substituting $v = \frac{1}{a}$ into the temperature equation: $T_{max} = \frac{p_{0}(\frac{1}{a})e^{-a(\frac{1}{a})}}{R} = \frac{p_{0}}{aeR}$.

For an ideal gas,the instantaneous change in pressure $p$ with volume $v$ is given by the equation $\frac{dp}{dv} = -ap$. If $p = p_{0}$ at $v = 0$ is the given boundary condition,then the maximum temperature one mole of gas can attain is: (Here $R$ is the gas constant)

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