One mole of an ideal gas undergoes a process | vedclass

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(B) Given the equation of state for the process: $P = 3 - g \left(\frac{V}{V_0}\right)^2$. Using the ideal gas law for $n = 1$ mole, $PV = RT$, we can

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$\frac{2 V_0}{R}$

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(B) Given the equation of state for the process: $P = 3 - g \left(\frac{V}{V_0}\right)^2$. Using the ideal gas law for $n = 1$ mole, $PV = RT$, we can write $T = \frac{PV}{R}$. Substituting $P$ in terms of $V$: $T = \frac{1}{R} \left[ 3V - g \frac{V^3}{V_0^2} \right]$. To find the maximum temperature, we differentiate $T$ with respect to $V$ and set it to zero: $\frac{dT}{dV} = \frac{1}{R} \left[ 3 - \frac{3gV^2}{V_0^2} \right] = 0$. This implies $3 = \frac{3gV^2}{V_0^2}$, so $V^2 = \frac{V_0^2}{g}$, or $V = \frac{V_0}{\sqrt{g}}$. Substituting this value of $V$ back into the expression for $T$: $T_{max} = \frac{1}{R} \left[ 3 \left(\frac{V_0}{\sqrt{g}}\right) - g \frac{(V_0/\sqrt{g})^3}{V_0^2} \right] = \frac{1}{R} \left[ \frac{3V_0}{\sqrt{g}} - \frac{V_0}{\sqrt{g}} \right] = \frac{2V_0}{R\sqrt{g}}$. Assuming the constant $g=1$ for standard dimensional consistency in the provided options, $T_{max} = \frac{2V_0}{R}$.

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