What work has to be done adiabatically to increase the root mean square speed of a mole of a diatomic gas `eta`=5 times from `T_1` = 300 K?

Answer

To solve the problem of finding the work done adiabatically to increase the root mean square (RMS) speed of a mole of a diatomic gas by 5 times from an initial temperature \( T_1 = 300 \, K \), we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Relationship Between RMS Speed and Temperature**:
The RMS speed \( V_{rms} \) of a gas is given by the formula:
\[ V_{rms} = \sqrt{\frac{3RT}{M}} \]
where \( R \) is the gas constant, \( T \) is the temperature in Kelvin, and \( M \) is the molar mass of the gas.

2. **Set Up the Initial and Final Conditions**:
Let the initial RMS speed at temperature \( T_1 \) be \( V_1 \) and the final RMS speed at temperature \( T_2 \) be \( V_2 \). According to the problem, we have:
\[ V_2 = 5V_1 \]

3. **Relate the RMS Speeds to the Temperatures**:
From the relationship of RMS speed and temperature, we can write:
\[ \frac{V_1}{V_2} = \sqrt{\frac{T_1}{T_2}} \]
Substituting \( V_2 = 5V_1 \):
\[ \frac{V_1}{5V_1} = \sqrt{\frac{T_1}{T_2}} \implies \frac{1}{5} = \sqrt{\frac{T_1}{T_2}} \]

4. **Square Both Sides to Eliminate the Square Root**:
Squaring both sides gives:
\[ \left(\frac{1}{5}\right)^2 = \frac{T_1}{T_2} \implies \frac{1}{25} = \frac{T_1}{T_2} \implies T_2 = 25T_1 \]

5. **Calculate the Final Temperature**:
Given \( T_1 = 300 \, K \):
\[ T_2 = 25 \times 300 \, K = 7500 \, K \]

6. **Calculate the Change in Internal Energy**:
For a diatomic gas, the degrees of freedom \( F = 5 \). The change in internal energy \( \Delta U \) is given by:
\[ \Delta U = \frac{F}{2} n R (T_2 - T_1) \]
For 1 mole of gas (\( n = 1 \)):
\[ \Delta U = \frac{5}{2} \times 1 \times R \times (T_2 - T_1) \]
Substituting \( R = 8.314 \, J/(mol \cdot K) \):
\[ \Delta U = \frac{5}{2} \times 8.314 \times (7500 - 300) \]
\[ \Delta U = \frac{5}{2} \times 8.314 \times 7200 \]

7. **Calculate the Work Done**:
In an adiabatic process, the work done \( W \) is given by:
\[ W = -\Delta U \]
Therefore:
\[ W = -\left(\frac{5}{2} \times 8.314 \times 7200\right) \]

8. **Final Calculation**:
Performing the calculation:
\[ W = -\left(\frac{5}{2} \times 8.314 \times 7200\right) \approx -1.49 \times 10^5 \, J \]

### Conclusion:
The work done adiabatically to increase the RMS speed of a mole of a diatomic gas by 5 times from \( T_1 = 300 \, K \) is approximately \( -1.49 \times 10^5 \, J \).