`8g` each of oxygen and hydrogen at `27^(@)C` will have the total kinetic energy in the ratio of …….

Explanation

To solve the problem of finding the ratio of total kinetic energy of oxygen to hydrogen, we can follow these steps:

### Step 1: Understand the formula for kinetic energy
The total kinetic energy (KE) of a gas can be expressed using the formula:
\[ KE = \frac{3}{2} nRT \]
where:
- \( n \) = number of moles of the gas
- \( R \) = gas constant
- \( T \) = temperature in Kelvin

### Step 2: Calculate the number of moles of each gas
We need to find the number of moles of oxygen (O₂) and hydrogen (H₂) using the formula:
\[ n = \frac{\text{mass (g)}}{\text{molar mass (g/mol)}} \]

**For Oxygen (O₂):**
- Mass = 8 g
- Molar mass of O₂ = 32 g/mol
\[ n_{O_2} = \frac{8 \text{ g}}{32 \text{ g/mol}} = \frac{1}{4} \text{ mol} \]

**For Hydrogen (H₂):**
- Mass = 8 g
- Molar mass of H₂ = 2 g/mol
\[ n_{H_2} = \frac{8 \text{ g}}{2 \text{ g/mol}} = 4 \text{ mol} \]

### Step 3: Set up the ratio of total kinetic energies
Now we can set up the ratio of the total kinetic energy of oxygen to that of hydrogen:
\[ \text{Ratio} = \frac{KE_{O_2}}{KE_{H_2}} = \frac{\frac{3}{2} n_{O_2} RT}{\frac{3}{2} n_{H_2} RT} \]

### Step 4: Simplify the ratio
Since \(\frac{3}{2}\) and \(RT\) are common in both the numerator and denominator, they cancel out:
\[ \text{Ratio} = \frac{n_{O_2}}{n_{H_2}} = \frac{\frac{1}{4}}{4} \]

### Step 5: Calculate the final ratio
Now we can calculate the final ratio:
\[ \text{Ratio} = \frac{1/4}{4} = \frac{1}{16} \]

### Final Answer
The total kinetic energy of oxygen and hydrogen will have the ratio of:
\[ \text{Ratio} = \frac{1}{16} \]

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