At what temperature is the `K.E`. Of a gas molecules half that of its value at `27^(@)C`

Options

A. `13.5^(@)C`
B. `150^(@)C`
C. `150 K`
D. `-123 K`

Explanation

To find the temperature at which the kinetic energy (K.E.) of gas molecules is half that of its value at \(27^\circ C\), we can follow these steps:

### Step 1: Understand the relationship between temperature and kinetic energy
The average kinetic energy of gas molecules is directly proportional to the absolute temperature (in Kelvin). The formula for the average kinetic energy \(U\) of a gas is given by:

\[ U = \frac{3}{2} NkT \]

where \(N\) is the number of molecules, \(k\) is the Boltzmann constant, and \(T\) is the absolute temperature in Kelvin.

### Step 2: Convert \(27^\circ C\) to Kelvin
To convert Celsius to Kelvin, we use the formula:

\[ T(K) = T(°C) + 273.15 \]

For \(27^\circ C\):

\[ T = 27 + 273.15 = 300.15 \, K \approx 300 \, K \]

### Step 3: Determine the kinetic energy at \(27^\circ C\)
At \(T = 300 \, K\), the average kinetic energy \(U\) is:

\[ U_{300} = \frac{3}{2} Nk(300) \]

### Step 4: Set up the equation for half the kinetic energy
We want to find the temperature \(T'\) at which the kinetic energy is half of \(U_{300}\):

\[ U_{T'} = \frac{1}{2} U_{300} \]

Substituting the expressions for kinetic energy, we have:

\[ \frac{3}{2} NkT' = \frac{1}{2} \left(\frac{3}{2} Nk(300)\right) \]

### Step 5: Simplify the equation
Cancelling out the common terms:

\[ T' = \frac{1}{2} \times 300 \]

### Step 6: Calculate \(T'\)
Now, calculating \(T'\):

\[ T' = 150 \, K \]

### Final Answer
The temperature at which the kinetic energy of the gas molecules is half that of its value at \(27^\circ C\) is:

\[ \boxed{150 \, K} \]

---