Question

The internal energy of a thermodynamic system is given by \( U = a s^{4/3} v^\alpha \), where \( s \) is entropy, \( v \) is volume, and \( a \) and \( \alpha \) are constants. The value of \( \alpha \) is:

• A. 1
• B. -1
• C. \(\frac{1}{3}\)
• D. \(-\frac{1}{3}\)

Hint:

In dimensional analysis, use the known dimensions of each physical quantity to determine the unknown constant α by ensuring dimensional consistency across the equation.

Answer 1

The Correct Option is D

Using thermodynamics and the Euler relation for extensive properties, consider \( U = as^{4/3}v^{\alpha} \).

The Euler relation is:

\[ U = Ts - Pv, \]

where \( T \) is temperature, \( s \) is entropy, \( P \) is pressure, and \( v \) is volume.

Given \( U = as^{4/3}v^{\alpha} \), we find the partial derivatives:

\[ T = \frac{\partial U}{\partial s} = \frac{4}{3}as^{1/3}v^{\alpha}, \]

\[ P = -\frac{\partial U}{\partial v} = -\alpha as^{4/3}v^{\alpha-1}. \]

Substitute into the Euler relation \( U = Ts - Pv \):

\[ as^{4/3}v^{\alpha} = \left( \frac{4}{3}as^{1/3}v^{\alpha} \right)s - \left( -\alpha as^{4/3}v^{\alpha-1} \right)v. \]

Simplify:

\[ as^{4/3}v^{\alpha} = \frac{4}{3}as^{4/3}v^{\alpha} + \alpha as^{4/3}v^{\alpha}. \]

Factor out \( as^{4/3}v^{\alpha} \):

\[ 1 = \frac{4}{3} + \alpha. \]

Solve for \( \alpha \):

\[ \alpha = 1 - \frac{4}{3} = -\frac{1}{3}. \]

Conclusion: Therefore, \( \alpha \) is:

\[ \boxed{-\frac{1}{3}} \]