Question

An ideal gas undergoes an adiabatic expansion from volume \( V \) to \( 2V \). If the initial temperature is \( T \), what is the final temperature? (Assume the ratio of specific heats \( \gamma = \frac{5}{3} \))

• A. \( T \)
• B. \( \frac{T}{2} \)
• C. \( \frac{T}{2^{2/3}} \)
• D. \( \frac{T}{2^{5/3}} \)

Hint:

\textbf{Tip:} For adiabatic processes, use \( TV^{\gamma - 1} = \text{const} \) to directly relate initial and final temperatures and volumes.

Answer 1

The Correct Option is C

The final temperature of an ideal gas during adiabatic expansion is determined using the adiabatic process relation:

\( TV^{\gamma-1} = \text{constant} \)

For an initial state ( \( T_1 = T \), \( V_1 = V \) ) and a final state ( \( T_2 \), \( V_2 = 2V \) ), the relation is:

\( TV^{\gamma-1} = T_2(2V)^{\gamma-1} \)

Simplification yields:

\( T V^{\gamma-1} = T_2 \cdot 2^{\gamma-1} \cdot V^{\gamma-1} \)

Canceling \( V^{\gamma-1} \) from both sides gives:

\( T = T_2 \cdot 2^{\gamma-1} \)

Solving for \( T_2 \):

\( T_2 = \frac{T}{2^{\gamma-1}} \)

Given \( \gamma = \frac{5}{3} \), calculate \( \gamma-1 \):

\( \gamma-1 = \frac{5}{3} - 1 = \frac{2}{3} \)

Substituting this value to find \( T_2 \):

\( T_2 = \frac{T}{2^{2/3}} \)

The final temperature, \( T_2 \), is \( \frac{T}{2^{2/3}} \).