Question

The ratio of adiabatic to the isobaric coefficient of expansion is

• A. \( \frac{\gamma - 1}{\gamma} \) (where \( \gamma = \frac{C_p}{C_v} \))
• B. \( \frac{1}{\gamma} \)
• C. \( \frac{1}{1 - \gamma} \)
• D. \( \frac{\gamma}{\gamma - 1} \)

Hint:

This result is related to the ratio of adiabatic compressibility (\(\kappa_S\)) to isothermal compressibility (\(\kappa_T\)), which is a more common relation: \( \frac{\kappa_S}{\kappa_T} = \frac{1}{\gamma} \). While the question asks about coefficients of expansion, the method involves similar derivations from the gas laws for different processes.

Answer 1

The Correct Option is C

Step 1: Understand the Objective:
The problem requires calculating the ratio of two coefficients of thermal expansion for a perfect gas: the adiabatic coefficient and the isobaric coefficient. The coefficient of volume expansion quantifies the change in volume of a substance with temperature variation.Step 2: Define Key Formulas:
The isobaric coefficient of volume expansion (at constant pressure) is given by:
\[ \alpha = \frac{1}{V} \left( \frac{\partial V}{\partial T} \right)_P \]The adiabatic coefficient of volume expansion (at constant entropy) is defined as:
\[ \beta = \frac{1}{V} \left( \frac{\partial V}{\partial T} \right)_S \]The goal is to determine the ratio \( \frac{\beta}{\alpha} \) for an ideal gas.Step 3: Derivations and Calculations:
1. Isobaric Coefficient (\(\alpha\)):
For one mole of a perfect gas, \( PV = RT \). At constant pressure, \( V = (R/P)T \).
The partial derivative is:
\[ \left( \frac{\partial V}{\partial T} \right)_P = \frac{R}{P} = \frac{V}{T} \]Therefore, the isobaric coefficient is:
\( \alpha = \frac{1}{V} \left( \frac{V}{T} \right) = \frac{1}{T} \).
2. Adiabatic Coefficient (\(\beta\)):
For an adiabatic process in a perfect gas, the relation is \( TV^{\gamma-1} = \text{constant} \). Differentiating with respect to T:
\[ d(TV^{\gamma-1}) = V^{\gamma-1}dT + T(\gamma-1)V^{\gamma-2}dV = 0 \]Solving for \( \frac{dV}{dT} \) under adiabatic conditions:
\[ T(\gamma-1)V^{\gamma-2}dV = -V^{\gamma-1}dT \]\[ \left( \frac{\partial V}{\partial T} \right)_S = -\frac{V^{\gamma-1}}{T(\gamma-1)V^{\gamma-2}} = -\frac{V}{T(\gamma-1)} \]The adiabatic coefficient is thus:
\( \beta = \frac{1}{V} \left( -\frac{V}{T(\gamma-1)} \right) = -\frac{1}{T(\gamma-1)} \).
3. Ratio Calculation (\( \frac{\beta}{\alpha} \)):
Compute the ratio of the adiabatic to the isobaric coefficient:
\[ \frac{\beta}{\alpha} = \frac{-1 / (T(\gamma-1))}{1 / T} = -\frac{1}{\gamma-1} \]This can be expressed as:
\[ \frac{1}{-(\gamma-1)} = \frac{1}{1 - \gamma} \]Step 4: Conclusion:
The ratio of the adiabatic to the isobaric coefficient of thermal expansion for a perfect gas is \( \frac{1}{1-\gamma} \).