An ideal gas is expanding such that PT^2 = te | vedclass

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(C) Given the process equation: $PT^2 = K$ (where $K$ is a constant).
Using the ideal gas law $PV = \mu RT$,we can express pressure as $P = \frac{\mu

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(C) Given the process equation: $PT^2 = K$ (where $K$ is a constant).
Using the ideal gas law $PV = \mu RT$,we can express pressure as $P = \frac{\mu RT}{V}$.
Substituting this into the process equation: $\left(\frac{\mu RT}{V}\right) T^2 = K$.
This simplifies to $\frac{\mu R T^3}{V} = K$,or $T^3 = \left(\frac{K}{\mu R}\right) V$.
Let $C = \frac{K}{\mu R}$,so $T^3 = CV$.
Differentiating both sides with respect to $T$: $3T^2 dT = C dV$.
Thus,$\frac{dV}{dT} = \frac{3T^2}{C} = \frac{3T^2}{(T^3/V)} = \frac{3V}{T}$.
The coefficient of volume expansion $\gamma$ is defined as $\gamma = \frac{1}{V} \frac{dV}{dT}$.
Substituting the expression for $\frac{dV}{dT}$: $\gamma = \frac{1}{V} \left(\frac{3V}{T}\right) = \frac{3}{T}$.

An ideal gas is expanding such that $PT^2 = \text{constant}$. The coefficient of volume expansion of the gas is (in $/T$)

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