Question

An ideal gas is expanding such that $pT^2$ = constant. The coefficient of volume expansion of the gas is

• A. $\frac{1}{T}$
• B. $\frac{2}{T}$
• C. $\frac{3}{T}$
• D. $\frac{4}{T}$

Answer 1

Answer: (C)

$\frac{3}{T}$

Answer 2

The coefficient of volume expansion $\beta$ is generally defined as $\beta = \frac{1}{V} \left(\frac{\partial V}{\partial T}\right)_p$. This is the isobaric coefficient of volume expansion. However, for a process defined by a specific relation between thermodynamic variables, the coefficient of volume expansion is often interpreted as $\frac{1}{V} \frac{dV}{dT}$ along the path of the process.

We are given that an ideal gas is expanding such that $pT^2 = C$, where $C$ is a constant. The equation of state for an ideal gas is $pV = nRT$, where $n$ is the number of moles and $R$ is the ideal gas constant.

From the ideal gas equation, we can express pressure $p$ as $p = \frac{nRT}{V}$. Substitute this expression for $p$ into the given process relation: $\left(\frac{nRT}{V}\right) T^2 = C$ $\frac{nRT^3}{V} = C$

Now, we can express the volume $V$ as a function of temperature $T$ for this process: $V = \frac{nRT^3}{C}$

To find the coefficient of volume expansion for this process, we need to calculate $\frac{1}{V} \frac{dV}{dT}$. First, let's find the derivative of $V$ with respect to $T$: $\frac{dV}{dT} = \frac{d}{dT}\left(\frac{nRT^3}{C}\right)$ Since $n$, $R$, and $C$ are constants, we can take them out of the differentiation: $\frac{dV}{dT} = \frac{nR}{C} \frac{d}{dT}(T^3)$ $\frac{dV}{dT} = \frac{nR}{C} (3T^2)$ $\frac{dV}{dT} = \frac{3nRT^2}{C}$

Now, we can calculate the coefficient of volume expansion $\beta_{process} = \frac{1}{V} \frac{dV}{dT}$: $\beta_{process} = \frac{1}{\left(\frac{nRT^3}{C}\right)} \left(\frac{3nRT^2}{C}\right)$ $\beta_{process} = \frac{C}{nRT^3} \times \frac{3nRT^2}{C}$

Cancel out the common terms $n$, $R$, $T^2$, and $C$: $\beta_{process} = \frac{3}{T}$

Thus, the coefficient of volume expansion of the gas for the given process $pT^2 = \text{constant}$ is $\frac{3}{T}$.