If the temperature of a gas is increased from `0^(@)C` to `273^(@)C`, then the ratio of average kinetic energy of the gas molecules is

Options

A. `1:4`
B. `4:1`
C. `1:1`
D. `2:1`

Explanation

To solve the problem of finding the ratio of the average kinetic energy of gas molecules when the temperature is increased from \(0^\circ C\) to \(273^\circ C\), we can follow these steps:

### Step 1: Convert the temperatures from Celsius to Kelvin
- The formula to convert Celsius to Kelvin is:
\[ T(K) = T(°C) + 273 \]
- For \(T_1 = 0^\circ C\):
\[ T_1 = 0 + 273 = 273 \, K \]
- For \(T_2 = 273^\circ C\):
\[ T_2 = 273 + 273 = 546 \, K \]

### Step 2: Understand the relationship between average kinetic energy and temperature
- The average kinetic energy (\(KE\)) of gas molecules is directly proportional to the absolute temperature \(T\). The formula for average kinetic energy is:
\[ KE \propto T \]
- Therefore, we can express the average kinetic energies at the two temperatures as:
\[ KE_1 \propto T_1 \quad \text{and} \quad KE_2 \propto T_2 \]

### Step 3: Set up the ratio of average kinetic energies
- The ratio of the average kinetic energies at the two temperatures can be expressed as:
\[ \frac{KE_2}{KE_1} = \frac{T_2}{T_1} \]

### Step 4: Substitute the values of \(T_1\) and \(T_2\)
- Now, substituting the values we found:
\[ \frac{KE_2}{KE_1} = \frac{546 \, K}{273 \, K} \]

### Step 5: Simplify the ratio
- Simplifying the ratio:
\[ \frac{KE_2}{KE_1} = \frac{546}{273} = 2 \]

### Conclusion
- Therefore, the ratio of the average kinetic energy of the gas molecules when the temperature is increased from \(0^\circ C\) to \(273^\circ C\) is:
\[ \frac{KE_2}{KE_1} = 2 \]