For an ideal gas at a temperature of 27^{circ | vedclass

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(D) For an ideal gas, the equation of state is $PV = nRT$.At constant pressure $P$, differentiating with respect to temperature $T$, we get $P \frac{d

Vedclass · Accepted Answer

$33 	imes 10^{-4} \ K^{-1}$

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(D) For an ideal gas, the equation of state is $PV = nRT$.At constant pressure $P$, differentiating with respect to temperature $T$, we get $P \frac{dV}{dT} = nR$, which implies $\frac{dV}{dT} = \frac{nR}{P}$.The coefficient of volume expansion $\gamma$ is defined as $\gamma = \frac{1}{V} \frac{dV}{dT}$.Substituting $\frac{dV}{dT} = \frac{nR}{P}$, we get $\gamma = \frac{1}{V} \left( \frac{nR}{P} ight) = \frac{nR}{PV}$.Since $PV = nRT$, we have $\gamma = \frac{nR}{nRT} = \frac{1}{T}$.Given temperature $T = 27^{\circ} C = 27 + 273 = 300 \ K$.Therefore, $\gamma = \frac{1}{300} \ K^{-1} \approx 0.00333 \ K^{-1} = 33 	imes 10^{-4} \ K^{-1}$.

For an ideal gas at a temperature of $27^{\circ} C$ and at constant pressure, the coefficient of volume expansion is nearly

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