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Chapter 17: Problem 66

Show that the coefficient of volume expansion of an ideal gas at constant pressure is the reciprocal of its kelvin temperature.

Short Answer

Expert verified
The coefficient of volume expansion (尾) at constant pressure for an ideal gas is equal to the reciprocal of the gas's absolute temperature. Hence \( 尾 = \frac{1}{T} \), with \( T \) as the temperature in Kelvins.

Step by step solution

01

Understanding Basic Principles

The first step is understanding the Ideal Gas Law, represented as \( PV = nRT \), where \( P \) is the pressure, \( V \) is the volume, \( n \) is the number of moles of the gas, \( R \) is the universal gas constant, and \( T \) is the temperature in kelvin. Considering the gas at constant pressure means that \( P \) is a constant. The relationship to be shown involves the the coefficient of volume expansion (尾), which by definition is \( 尾 = \frac{1}{V}\frac{dV}{dT} \). This defines how the volume of a substance changes with temperature, with pressure held constant.
02

Derivation of the Relationship

Next, substitute \( V = \frac{nRT}{P} \) from the Ideal Gas Law into the equation for 尾. This gives \( 尾 = \frac{1}{\frac{nRT}{P}} \frac{d(\frac{nRT}{P})}{dT} \). Then find the derivative with respect to \( T \) to get \( 尾 = \frac{P}{nRT} \times nR.P^{-1} \). Simplifying this, cancelling out \( P \) and \( nR \), yields \( 尾 = \frac{1}{T} \). This shows that the volume expansion coefficient (尾) of an ideal gas at constant pressure is indeed the reciprocal of its absolute temperature \( T \) in kelvin.

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