A vessel containing one gram -mole of oxygen is enclosed in a thermally insulated vessel. The vessel is next moved with a constant speed `v_(0)` and then suddenly stopped. The process results in a rise in the temperature of the gas by `1^(@)c`. Calculate the speed `v_(0)`.

Explanation

To solve the problem, we need to determine the speed \( v_0 \) of a vessel containing one gram-mole of oxygen that results in a temperature rise of \( 1^\circ C \) when the vessel is suddenly stopped.

### Step-by-Step Solution:

1. **Understanding the Problem**:
- We have 1 gram-mole of oxygen in a thermally insulated vessel.
- The vessel is moving with a constant speed \( v_0 \) and is then suddenly stopped.
- This stopping leads to a temperature rise of \( 1^\circ C \).

2. **Using the First Law of Thermodynamics**:
- In a thermally insulated system, the change in internal energy (\( \Delta U \)) is equal to the work done on the system.
- The work done on the gas when the vessel is stopped can be equated to the change in internal energy, which is related to the temperature change.

3. **Kinetic Energy**:
- The kinetic energy (KE) of the gas when moving at speed \( v_0 \) is given by:
\[ KE = \frac{1}{2} mv_0^2 \]
- Here, \( m \) is the mass of the gas.

4. **Change in Internal Energy**:
- The change in internal energy due to the temperature change can be expressed as:
\[ \Delta U = n C_v \Delta T \]
- Where \( n \) is the number of moles, \( C_v \) is the specific heat at constant volume, and \( \Delta T \) is the change in temperature.

5. **Specific Heat for Oxygen**:
- For one mole of oxygen, \( n = 1 \).
- The specific heat at constant volume \( C_v \) for a diatomic gas like oxygen is given by:
\[ C_v = \frac{R}{\gamma - 1} \]
- Where \( R \) is the universal gas constant \( (R = 8.314 \, \text{J/(mol K)}) \) and \( \gamma \) (the ratio of specific heats) for oxygen is approximately \( 1.41 \).

6. **Substituting Values**:
- The temperature change \( \Delta T = 1^\circ C = 1 \, \text{K} \).
- Substitute \( C_v \) into the internal energy equation:
\[ \Delta U = 1 \cdot \frac{R}{\gamma - 1} \cdot 1 \]
- Thus,
\[ \Delta U = \frac{R}{\gamma - 1} \]

7. **Equating Kinetic Energy and Change in Internal Energy**:
- Set the kinetic energy equal to the change in internal energy:
\[ \frac{1}{2} mv_0^2 = \frac{R}{\gamma - 1} \]

8. **Mass of Oxygen**:
- The molar mass of oxygen is \( 32 \, \text{g/mol} \), which is \( 0.032 \, \text{kg} \) for calculations.

9. **Rearranging for \( v_0 \)**:
- Rearranging gives:
\[ v_0^2 = \frac{2R}{\gamma - 1} \cdot \frac{1}{m} \]
- Therefore,
\[ v_0 = \sqrt{\frac{2R}{\gamma - 1} \cdot \frac{1}{m}} \]

10. **Substituting Known Values**:
- Substitute \( R = 8.314 \, \text{J/(mol K)} \), \( \gamma = 1.41 \), and \( m = 0.032 \, \text{kg} \):
\[ v_0 = \sqrt{\frac{2 \cdot 8.314}{1.41 - 1} \cdot \frac{1}{0.032}} \]

11. **Calculating**:
- Calculate the expression:
\[ v_0 = \sqrt{\frac{16.628}{0.41} \cdot 31.25} \]
\[ v_0 = \sqrt{16.628 \cdot 76.56} \approx \sqrt{1278.56} \approx 35.6 \, \text{m/s} \]

### Final Answer:
The speed \( v_0 \) is approximately \( 35.6 \, \text{m/s} \).