Question

$\gamma_A$ is the specific heat ratio of monoatomic gas A having 3 translational degrees of freedom. $\gamma_B$ is the specific heat ratio of polyatomic gas B having 3 translational, 3 rotational degrees of freedom and 1 vibrational mode. If \[ \frac{\gamma_A}{\gamma_B} = \left( 1 + \frac{1}{n} \right) \] then the value of \( n \) is ___.

Hint:

For problems involving specific heat ratios, break down the equation and substitute values for the translational, rotational, and vibrational degrees of freedom separately. This will help you arrive at the correct value of $n$.

Answer 1

Correct Answer: 3

\[ \frac{\gamma_A}{\gamma_B} = \frac{f_A + 2}{f_A} : \text{for monoatomic gas A} \] \[ \frac{\gamma_B}{\gamma_B} = \frac{f_B + 2}{f_B} : \text{for polyatomic gas B} \] For monoatomic gas A: \[ f_A = 3 \quad \text{(translational degrees of freedom)} \] For polyatomic gas B: \[ f_B = 3 + 3 + 1 = 7 \quad \text{(translational, rotational, and vibrational modes)} \] Substituting these values into the formula: \[ \frac{\gamma_A}{\gamma_B} = \frac{3+2}{3} : \frac{7+2}{7} = \frac{5}{3} : \frac{9}{7} \] \[ \frac{5}{3} : \frac{9}{7} = \left( 1 + \frac{1}{n} \right) \] \[ \frac{5}{3} \cdot \frac{7}{9} = 1 + \frac{1}{n} \] \[ \frac{35}{27} = 1 + \frac{1}{n} \] \[ \frac{35}{27} - 1 = \frac{1}{n} \] \[ \frac{8}{27} = \frac{1}{n} \Rightarrow n = 3 \]

Answer 2

The problem asks to find the value of \( n \) from the given relation \( \frac{\gamma_A}{\gamma_B} = \left( 1 + \frac{1}{n} \right) \), where \( \gamma_A \) and \( \gamma_B \) are the specific heat ratios for a monoatomic gas A and a polyatomic gas B, respectively, with their degrees of freedom specified.

Concept Used:

The specific heat ratio (or adiabatic index), \( \gamma \), of a gas is related to its total degrees of freedom, \( f \), by the formula:

\[ \gamma = 1 + \frac{2}{f} \]

The total degrees of freedom (\( f \)) is the sum of translational, rotational, and vibrational degrees of freedom.

• Each translational degree of freedom contributes \( \frac{1}{2}kT \) to the internal energy.
• Each rotational degree of freedom contributes \( \frac{1}{2}kT \) to the internal energy.
• Each vibrational mode contributes \( kT \) to the internal energy ( \( \frac{1}{2}kT \) for kinetic and \( \frac{1}{2}kT \) for potential energy), which corresponds to 2 degrees of freedom.

Step-by-Step Solution:

Step 1: Calculate the degrees of freedom for the monoatomic gas A (\( f_A \)).

A monoatomic gas has only translational motion. It is given that there are 3 translational degrees of freedom.

\[ f_A = f_{\text{trans}} = 3 \]

Step 2: Calculate the specific heat ratio for gas A (\( \gamma_A \)).

Using the formula for \( \gamma \):

\[ \gamma_A = 1 + \frac{2}{f_A} = 1 + \frac{2}{3} = \frac{3+2}{3} = \frac{5}{3} \]

Step 3: Calculate the degrees of freedom for the polyatomic gas B (\( f_B \)).

Gas B has 3 translational, 3 rotational, and 1 vibrational mode. As each vibrational mode corresponds to 2 degrees of freedom:

\[ f_B = f_{\text{trans}} + f_{\text{rot}} + f_{\text{vib}} \] \[ f_B = 3 + 3 + (1 \times 2) = 8 \]

Step 4: Calculate the specific heat ratio for gas B (\( \gamma_B \)).

Using the formula for \( \gamma \):

\[ \gamma_B = 1 + \frac{2}{f_B} = 1 + \frac{2}{8} = 1 + \frac{1}{4} = \frac{4+1}{4} = \frac{5}{4} \]

Step 5: Use the given relation to find the value of \( n \).

The relation is:

\[ \frac{\gamma_A}{\gamma_B} = 1 + \frac{1}{n} \]

Substitute the calculated values of \( \gamma_A \) and \( \gamma_B \):

\[ \frac{5/3}{5/4} = 1 + \frac{1}{n} \] \[ \frac{5}{3} \times \frac{4}{5} = 1 + \frac{1}{n} \] \[ \frac{4}{3} = 1 + \frac{1}{n} \]

Step 6: Solve for \( n \).

\[ \frac{1}{n} = \frac{4}{3} - 1 \] \[ \frac{1}{n} = \frac{4 - 3}{3} = \frac{1}{3} \] \[ n = 3 \]

The value of \( n \) is 3.