Question

The specific heat capacity at constant volume \(C_v\) of a mole of an ideal gas is related to the gas constant \(R\) and the ratio of the specific heats \(\gamma\) as

• A. \(\frac{R}{\gamma-1}\)
• B. \(\frac{R\gamma}{1-\gamma}\)
• C. \(\frac{R(1-\gamma)}{\gamma}\)
• D. \(\frac{R}{1-\gamma}\)
• E. \(\frac{R}{1+\gamma}\)

Hint:

Always remember the two key relations: \(C_p-C_v=R\) and \(\gamma=\frac{C_p}{C_v}\). These are enough to derive any required formula.

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
This question requires knowledge of the relationship between the molar specific heat at constant pressure (\(C_p\)), molar specific heat at constant volume (\(C_v\)), the universal gas constant (\(R\)), and the adiabatic index or ratio of specific heats (\(\gamma\)).
Step 2: Key Formula or Approach:
There are two fundamental relations for an ideal gas that we will use:
1. Mayer's Relation: \(C_p - C_v = R\)
2. Definition of Adiabatic Index (\(\gamma\)): \(\gamma = \frac{C_p}{C_v}\)
We need to combine these two equations to express \(C_v\) in terms of \(R\) and \(\gamma\).
Step 3: Detailed Explanation:
From the definition of \(\gamma\), we can write \(C_p\) in terms of \(C_v\):
\[ C_p = \gamma C_v \] Now, substitute this expression for \(C_p\) into Mayer's relation:
\[ C_p - C_v = R \] \[ (\gamma C_v) - C_v = R \] Factor out \(C_v\) from the left side of the equation:
\[ C_v (\gamma - 1) = R \] Finally, to solve for \(C_v\), divide both sides by \((\gamma - 1)\):
\[ C_v = \frac{R}{\gamma - 1} \] Step 4: Final Answer:
The specific heat capacity at constant volume, \(C_v\), is given by the expression \(\frac{R}{\gamma - 1}\). This matches option (A).