The volume of a diatomic gas `(gamma = 7//5)` is increased two times in a polytropic process with molar heat capacity ` C = R`. How many times will the rate of collision of molecules against the wall of the vessel be reduced as a result of this process?

Explanation

To solve the problem, we need to determine how many times the rate of collision of molecules against the wall of the vessel is reduced when the volume of a diatomic gas is increased two times in a polytropic process with a given molar heat capacity.

### Step-by-Step Solution:

1. **Identify Given Values:**
- For a diatomic gas, \(\gamma = \frac{7}{5}\).
- Molar heat capacity \(C = R\).
- The volume is increased two times: \(V_f = 2V_i\).

2. **Determine the Polytropic Process Relationship:**
- In a polytropic process, the relationship is given by \(PV^x = \text{constant}\).
- We need to find the value of \(x\) using the heat capacity relation:
\[ C = \frac{R}{1 - x} + \frac{R}{\gamma - 1} \]
- Substituting \(C = R\) and \(\gamma = \frac{7}{5}\):
\[ R = \frac{R}{1 - x} + \frac{R}{\frac{7}{5} - 1} \]
- Simplifying gives:
\[ R = \frac{R}{1 - x} + \frac{5R}{2} \]
- Rearranging leads to:
\[ 1 = \frac{1}{1 - x} + \frac{5}{2} \]
- This simplifies to:
\[ 1 - x = \frac{2}{3} \implies x = \frac{5}{3} \]

3. **Find the Final Pressure:**
- Using the polytropic relation \(PV^{\frac{5}{3}} = \text{constant}\):
- If \(V_f = 2V_i\), then:
\[ P_f (2V_i)^{\frac{5}{3}} = P_i V_i^{\frac{5}{3}} \]
- Rearranging gives:
\[ P_f = P_i \left(\frac{V_i^{\frac{5}{3}}}{(2V_i)^{\frac{5}{3}}}\right) = P_i \left(\frac{1}{2^{\frac{5}{3}}}\right) \]

4. **Calculate the Rate of Collision:**
- The rate of collision \(Z\) is proportional to the pressure \(P\) and the root mean square speed \(V_{rms}\):
\[ Z \propto \frac{P}{V_{rms}} \]
- The \(V_{rms}\) is given by:
\[ V_{rms} \propto \sqrt{\frac{P}{V}} \]
- Therefore, we can express the new \(V_{rms}\):
\[ V_{rms,f} = V_{rms,i} \cdot \sqrt{\frac{P_f}{P_i}} \cdot \sqrt{\frac{V_i}{2V_i}} = V_{rms,i} \cdot \sqrt{\frac{1}{2^{\frac{5}{3}}}} \cdot \sqrt{\frac{1}{2}} = V_{rms,i} \cdot \frac{1}{2^{\frac{5}{6}}} \]

5. **Final Rate of Collision Calculation:**
- Substituting \(P_f\) and \(V_{rms,f}\) into the collision rate equation:
\[ Z_f \propto \frac{P_f}{V_{rms,f}} = \frac{P_i \cdot \frac{1}{2^{\frac{5}{3}}}}{V_{rms,i} \cdot \frac{1}{2^{\frac{5}{6}}}} = Z_i \cdot \frac{1}{2^{\frac{5}{3}}} \cdot 2^{\frac{5}{6}} = Z_i \cdot 2^{-\frac{5}{3} + \frac{5}{6}} = Z_i \cdot 2^{-\frac{10}{6} + \frac{5}{6}} = Z_i \cdot 2^{-\frac{5}{6}} \]

6. **Conclusion:**
- The rate of collision is reduced by a factor of \(2^{\frac{5}{6}}\).

### Final Answer:
The rate of collision of molecules against the wall of the vessel is reduced by a factor of \(2^{\frac{5}{6}}\).