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

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(C) The coefficient of volume expansion is defined as $\gamma = \frac{1}{V} \left( \frac{dV}{dT} ight)$.Given the process equation: $PT^2 = 	ext{co

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$\frac{3}{T}$

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(C) The coefficient of volume expansion is defined as $\gamma = \frac{1}{V} \left( \frac{dV}{dT} ight)$.Given the process equation: $PT^2 = 	ext{constant}$.Using the ideal gas law $PV = nRT$, we can write $P = \frac{nRT}{V}$.Substituting $P$ into the process equation: $\left( \frac{nRT}{V} ight) T^2 = 	ext{constant}$.This simplifies to $\frac{T^3}{V} = 	ext{constant}$, or $V = k T^3$ (where $k$ is a constant).Differentiating $V$ with respect to $T$: $\frac{dV}{dT} = 3k T^2$.Now, substitute these into the expression for $\gamma$:$\gamma = \frac{1}{V} \left( \frac{dV}{dT} ight) = \frac{1}{kT^3} (3kT^2) = \frac{3}{T}$.

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

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