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Animated Solution for Physics - Thermodynamics: A rigid diatomic ideal gas undergoes an adiabatic process at room temperature. The relation between temperature and volume for this process is , then is

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Visualized Solution

Visualizing the System

  • A rigid diatomic ideal gas undergoes an adiabatic process.

Degree of Freedom ()

  • For a rigid diatomic gas at room temperature:

Adiabatic Index ()

Standard Adiabatic Relation

  • For an adiabatic process, the relation between temperature and volume is:

Comparing Equations

  • Given relation:
  • Comparing the powers of :

Calculating

Final Answer

The Sigma Insight: Thermodynamic Processes

Solution Diagram

Unraveling the Adiabatic Mystery

Finding the Power of Volume
Imagine a perfectly insulated cylinder containing a rigid diatomic gas. It is undergoing an adiabatic process, meaning absolutely no heat enters or leaves the system. Our goal is to find the exact relationship between its temperature and volume during this rapid expansion or compression.

The Degree of Freedom

Before we dive into the thermodynamics, we need to understand the nature of our gas. The problem specifies it is a rigid diatomic gas at room temperature.
What does this mean for its degrees of freedom? A diatomic molecule, like a tiny dumbbell, can move in three independent directions (translational) and rotate about two independent axes (rotational). Because it is rigid, the atoms don't vibrate along the bond. This gives us a total degree of freedom:
With the degree of freedom known, we can find the adiabatic index, (gamma). The formula connecting them is:
Substituting , we get:

The Master Equation

Next, let's recall the standard equation relating temperature () and volume () for an adiabatic process. Derived from the ideal gas law and the adiabatic condition , the relation is:

The Final Deduction

The problem gives us a specific relation for this process:
By comparing our standard equation with the given one, we can clearly see that the exponent must be equal to . Let's set up the equation:
Now, we simply plug in the value of we found earlier:
Taking the lowest common multiple to subtract the fractions:
And that simplifies beautifully to . This is our required value for , proving that even abstract thermodynamic relations boil down to simple, elegant fractions when you understand the underlying physics.

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