Question

An ideal gas undergoes a thermodynamic process following the relation PT² = constant. Assuming symbols have their usual meaning then volume expansion coefficient of the gas is equal to

• A. \(\frac{2}{T}\)
• B. \(\frac{3}{T}\)
• C. \(\frac{1}{2T}\)
• D. \(\frac{1}{T}\)

Answer 1

The Correct Option is B

To determine the volume expansion coefficient of the gas given the relation \( PT^2 = \text{constant} \), we start by analyzing the ideal gas law and the provided relation.

According to the ideal gas law, the state of an ideal gas is given by:

PV = nRT

where \( P \) is pressure, \( V \) is volume, \( n \) is number of moles, \( R \) is the universal gas constant, and \( T \) is temperature in Kelvin.

The given thermodynamic process is characterized by the relation \( PT^2 = \text{constant} \), let's denote this constant as \( C \):

P T^2 = C \quad \Rightarrow \quad P = \frac{C}{T^2}

Substitute this expression for \( P \) into the ideal gas law:

\frac{C}{T^2} \cdot V = nRT \quad \Rightarrow \quad V = \frac{nRT^3}{C}

The volume expansion coefficient, \( \gamma \), is defined as:

\gamma = \frac{1}{V} \left( \frac{\partial V}{\partial T} \right)_P

Now differentiate the expression for \( V \) with respect to \( T \) while keeping \( P \) constant:

The differential of \( V = \frac{nRT^3}{C} \) with respect to \( T \) gives:

\frac{\partial V}{\partial T} = \frac{3nRT^2}{C}

Now substitute back into the expression for \( \gamma \):

\gamma = \frac{1}{V} \left( \frac{\partial V}{\partial T} \right)_P = \frac{C}{nRT^3} \cdot \frac{3nRT^2}{C}

Simplifying this expression, we find:

\gamma = \frac{3}{T}

Thus, the volume expansion coefficient of the gas is \frac{3}{T}, which verifies that the correct option is indeed \frac{3}{T}.