A vessel containing one gram-mole of oxygen is enclosed in a thermally insulated vessel. The vessel is than 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`.

Answer

To solve the problem, we need to calculate the speed \( v_0 \) of the vessel containing one gram-mole of oxygen gas that results in a temperature rise of \( 1^\circ C \) when the vessel is suddenly stopped. Here's the step-by-step solution:

### Step 1: Understand the energy conversion
When the vessel is moving with speed \( v_0 \), it possesses kinetic energy. Upon stopping, this kinetic energy is converted into internal energy of the gas, which results in a temperature increase.

### Step 2: Write the expression for kinetic energy
The kinetic energy \( KE \) of the moving vessel is given by:
\[ KE = \frac{1}{2} m v_0^2 \]
where \( m \) is the mass of the gas.

For one mole of oxygen, the molar mass \( M \) is approximately \( 32 \, g/mol \) (since oxygen is \( O_2 \)). Thus, the mass \( m \) in kilograms is:
\[ m = 0.032 \, kg \]

So the kinetic energy can be expressed as:
\[ KE = \frac{1}{2} \times 0.032 \times v_0^2 \]

### Step 3: Write the expression for change in internal energy
The change in internal energy \( \Delta U \) of the gas can be expressed using the formula:
\[ \Delta U = n C_v \Delta T \]
where:
- \( n \) is the number of moles (1 mole),
- \( C_v \) is the molar specific heat at constant volume for oxygen, which is \( \frac{5}{2} R \) (where \( R \) is the universal gas constant, approximately \( 8.314 \, J/(mol \cdot K) \)),
- \( \Delta T \) is the change in temperature (given as \( 1^\circ C = 1 \, K \)).

Substituting the values:
\[ \Delta U = 1 \times \frac{5}{2} R \times 1 \]
\[ \Delta U = \frac{5}{2} R \]

### Step 4: Set the kinetic energy equal to the change in internal energy
Setting the kinetic energy equal to the change in internal energy gives:
\[ \frac{1}{2} \times 0.032 \times v_0^2 = \frac{5}{2} R \]

### Step 5: Substitute the value of \( R \)
Substituting \( R = 8.314 \, J/(mol \cdot K) \):
\[ \frac{1}{2} \times 0.032 \times v_0^2 = \frac{5}{2} \times 8.314 \]
\[ 0.016 v_0^2 = 20.785 \]

### Step 6: Solve for \( v_0^2 \)
\[ v_0^2 = \frac{20.785}{0.016} \]
\[ v_0^2 = 1292.8125 \]

### Step 7: Calculate \( v_0 \)
Taking the square root:
\[ v_0 = \sqrt{1292.8125} \approx 36.0 \, m/s \]

### Final Answer
The speed \( v_0 \) is approximately \( 36.0 \, m/s \).

---