Calculating Specific Heat Ratio Based on Degrees of Freedom

The specific heat ratio, denoted by $\gamma$, relates the heat capacities of a gas at constant pressure ($C_p$) and constant volume ($C_v$). For an ideal gas, it depends on the degrees of freedom ($f$) according to the formula:

$ \gamma = 1 + \frac{2}{f} $

Specific Heat Ratio for Monoatomic Gas A ($\gamma_A$)

Gas A is monoatomic and has 3 translational degrees of freedom ($f_A = 3$).

Using the formula:

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

Specific Heat Ratio for Polyatomic Gas B ($\gamma_B$)

Gas B is polyatomic with:

• 3 translational degrees of freedom
• 3 rotational degrees of freedom
• 1 vibrational mode

Each mode (translational, rotational, vibrational) typically contributes 2 to the effective degrees of freedom in the context of energy calculation ($f = f_{trans} + f_{rot} + f_{vib}$).

Total degrees of freedom for Gas B:

$ f_B = 3 \text{ (trans)} + 3 \text{ (rot)} + 2 \text{ (vib)} = 8 $

Using the formula:

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

Finding the Value of n

We are given the relation:

$ \frac{\gamma_A}{\gamma_B} = \left(1 + \frac{1}{n}\right) $

First, calculate the ratio $\frac{\gamma_A}{\gamma_B}$:

$ \frac{\gamma_A}{\gamma_B} = \frac{5/3}{5/4} $

$ \frac{\gamma_A}{\gamma_B} = \frac{5}{3} \times \frac{4}{5} = \frac{4}{3} $

Now, substitute this ratio back into the given relation:

$ \frac{4}{3} = 1 + \frac{1}{n} $

Solve for $\frac{1}{n}$:

$ \frac{1}{n} = \frac{4}{3} - 1 = \frac{4-3}{3} = \frac{1}{3} $

Therefore, the value of n is:

$ n = 3 $