If `T_(1),T_(2)` and `T_(3)` are the temperatures at which the `u_("rms"),u_("average"),u_("mp")` of oxygen gas are all equal to 1500 m/s then the correct statement is :

Options

A. `T_(1)gtT_(2)gtT_(3)`
B. `T_(1)ltT_(2)ltT_(3)`
C. `T_(1)=T_(2)=T_(3)`
D. None of these

Explanation

To solve the problem, we need to find the temperatures \( T_1, T_2, \) and \( T_3 \) at which the root mean square speed (\( u_{rms} \)), average speed (\( u_{average} \)), and most probable speed (\( u_{mp} \)) of oxygen gas are all equal to 1500 m/s. We will use the formulas for each of these speeds and derive the corresponding temperatures.

### Step 1: Write the formula for \( u_{rms} \)

The formula for the root mean square speed (\( u_{rms} \)) is given by:

\[ u_{rms} = \sqrt{\frac{3RT}{M}} \]

Where:
- \( R \) is the universal gas constant,
- \( T \) is the temperature in Kelvin,
- \( M \) is the molar mass of the gas.

Given that \( u_{rms} = 1500 \, \text{m/s} \), we can set up the equation:

\[ \sqrt{\frac{3RT_1}{M}} = 1500 \]

### Step 2: Solve for \( T_1 \)

Squaring both sides gives:

\[ \frac{3RT_1}{M} = 1500^2 \]

Rearranging for \( T_1 \):

\[ T_1 = \frac{1500^2 \cdot M}{3R} \]

Calculating \( T_1 \):

Assuming \( R \) and \( M \) are constants, we can compute \( T_1 \) numerically. Let's say \( R = 8.314 \, \text{J/(mol·K)} \) and \( M \) for oxygen is approximately \( 32 \, \text{g/mol} = 0.032 \, \text{kg/mol} \):

\[ T_1 = \frac{1500^2 \cdot 0.032}{3 \cdot 8.314} \]

Calculating this gives:

\[ T_1 \approx 750000 \, \text{K} \]

### Step 3: Write the formula for \( u_{average} \)

The formula for the average speed (\( u_{average} \)) is given by:

\[ u_{average} = \sqrt{\frac{8RT}{\pi M}} \]

Setting \( u_{average} = 1500 \, \text{m/s} \):

\[ \sqrt{\frac{8RT_2}{\pi M}} = 1500 \]

### Step 4: Solve for \( T_2 \)

Squaring both sides gives:

\[ \frac{8RT_2}{\pi M} = 1500^2 \]

Rearranging for \( T_2 \):

\[ T_2 = \frac{1500^2 \cdot \pi M}{8R} \]

Calculating \( T_2 \):

Using the same values for \( R \) and \( M \):

\[ T_2 = \frac{1500^2 \cdot \pi \cdot 0.032}{8 \cdot 8.314} \]

Calculating this gives:

\[ T_2 \approx 883125 \, \text{K} \]

### Step 5: Write the formula for \( u_{mp} \)

The formula for the most probable speed (\( u_{mp} \)) is given by:

\[ u_{mp} = \sqrt{\frac{2RT}{M}} \]

Setting \( u_{mp} = 1500 \, \text{m/s} \):

\[ \sqrt{\frac{2RT_3}{M}} = 1500 \]

### Step 6: Solve for \( T_3 \)

Squaring both sides gives:

\[ \frac{2RT_3}{M} = 1500^2 \]

Rearranging for \( T_3 \):

\[ T_3 = \frac{1500^2 \cdot M}{2R} \]

Calculating \( T_3 \):

Using the same values for \( R \) and \( M \):

\[ T_3 = \frac{1500^2 \cdot 0.032}{2 \cdot 8.314} \]

Calculating this gives:

\[ T_3 \approx 1125000 \, \text{K} \]

### Step 7: Compare the temperatures

Now we have:

- \( T_1 \approx 750000 \, \text{K} \)
- \( T_2 \approx 883125 \, \text{K} \)
- \( T_3 \approx 1125000 \, \text{K} \)

From this, we can conclude:

\[ T_1 < T_2 < T_3 \]

### Final Answer

The correct statement is \( T_1 < T_2 < T_3 \).