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: Understanding the Concept:
The question asks for the ratio of two coefficients of thermal expansion: the adiabatic coefficient and the isobaric coefficient. The coefficient of volume expansion describes how the volume of a substance changes with a change in temperature.
Step 2: Key Formula or Approach:
The isobaric (constant pressure) coefficient of volume expansion is defined as:
\[ \alpha = \frac{1}{V} \left( \frac{\partial V}{\partial T} \right)_P \] The adiabatic (constant entropy) coefficient of volume expansion is defined as:
\[ \beta = \frac{1}{V} \left( \frac{\partial V}{\partial T} \right)_S \] We need to find the ratio \( \frac{\beta}{\alpha} \) for a perfect gas.
Step 3: Detailed Explanation:
1. Calculate the isobaric coefficient (\(\alpha\)):
For one mole of a perfect gas, \( PV = RT \). At constant pressure, \( V = (R/P)T \).
\[ \left( \frac{\partial V}{\partial T} \right)_P = \frac{R}{P} = \frac{V}{T} \] So, \( \alpha = \frac{1}{V} \left( \frac{V}{T} \right) = \frac{1}{T} \).
2. Calculate the adiabatic coefficient (\(\beta\)):
For an adiabatic process, \( TV^{\gamma-1} = \text{constant} \). Differentiating this with respect to T:
\[ d(TV^{\gamma-1}) = V^{\gamma-1}dT + T(\gamma-1)V^{\gamma-2}dV = 0 \] Rearranging to find \( \frac{dV}{dT} \) for this process:
\[ T(\gamma-1)V^{\gamma-2}dV = -V^{\gamma-1}dT \] \[ \left( \frac{\partial V}{\partial T} \right)_S = -\frac{V^{\gamma-1}}{T(\gamma-1)V^{\gamma-2}} = -\frac{V}{T(\gamma-1)} \] So, \( \beta = \frac{1}{V} \left( -\frac{V}{T(\gamma-1)} \right) = -\frac{1}{T(\gamma-1)} \).
3. Find the ratio \( \frac{\beta}{\alpha} \):
\[ \frac{\text{Adiabatic coefficient}}{\text{Isobaric coefficient}} = \frac{\beta}{\alpha} = \frac{-1 / (T(\gamma-1))}{1 / T} = -\frac{1}{\gamma-1} \] This can be rewritten as:
\[ \frac{1}{-(\gamma-1)} = \frac{1}{1 - \gamma} \] Step 4: Final Answer:
The ratio of the adiabatic to the isobaric coefficient of expansion is \( \frac{1}{1-\gamma} \).