The pressure (P) and temperature (T) relation | vedclass

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(D) Given the relation $PT^2 = C$ (where $C$ is a constant).Using the ideal gas law $PV = nRT$, we can write $P = \frac{nRT}{V}$.Substituting $P$ into

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$\frac{3}{T}$

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(D) Given the relation $PT^2 = C$ (where $C$ is a constant).Using the ideal gas law $PV = nRT$, we can write $P = \frac{nRT}{V}$.Substituting $P$ into the given equation: $\left(\frac{nRT}{V}ight)T^2 = C$.This simplifies to $\frac{nRT^3}{V} = C$, which implies $V = \left(\frac{nR}{C}ight)T^3$.Let $K = \frac{nR}{C}$, so $V = KT^3$.The volume expansion coefficient $\gamma$ is defined as $\gamma = \frac{1}{V} \frac{dV}{dT}$.Differentiating $V$ with respect to $T$: $\frac{dV}{dT} = 3KT^2$.Now, substitute into the formula for $\gamma$: $\gamma = \frac{1}{KT^3} 	imes (3KT^2) = \frac{3}{T}$.

The pressure $(P)$ and temperature $(T)$ relationship of an ideal gas obeys the equation $PT^2 = \text{constant}$. The volume expansion coefficient of the gas will be:

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