Temperature at which the kinetic energy of gas molecule is half of the value of kinetic energy at `27^(@) ` C is

Options

A. `13.5^(@)` C
B. `150^(@)` C
C. 75 K
D. `-123^(@) C `

Explanation

To find the temperature at which the kinetic energy of gas molecules is half of the kinetic energy at 27°C, we can follow these steps:

### Step 1: Understand the relationship between kinetic energy and temperature
The kinetic energy (KE) of a gas molecule is given by the formula:
\[ KE = \frac{3}{2} k T \]
where \( k \) is the Boltzmann constant and \( T \) is the absolute temperature in Kelvin.

### Step 2: Convert the given temperature to Kelvin
The given temperature is \( 27^\circ C \). To convert this to Kelvin:
\[ T_1 = 27 + 273 = 300 \, K \]

### Step 3: Set up the equation for kinetic energy
Let the kinetic energy at \( 27^\circ C \) be \( KE_1 \):
\[ KE_1 = \frac{3}{2} k T_1 = \frac{3}{2} k \times 300 \]

Now, we want to find the temperature \( T_2 \) at which the kinetic energy \( KE_2 \) is half of \( KE_1 \):
\[ KE_2 = \frac{1}{2} KE_1 = \frac{1}{2} \left( \frac{3}{2} k \times 300 \right) = \frac{3}{4} k \times 300 \]

### Step 4: Express \( KE_2 \) in terms of \( T_2 \)
Using the kinetic energy formula for \( T_2 \):
\[ KE_2 = \frac{3}{2} k T_2 \]

### Step 5: Set the equations equal to each other
Now we can set the two expressions for kinetic energy equal to each other:
\[ \frac{3}{2} k T_2 = \frac{3}{4} k \times 300 \]

### Step 6: Simplify the equation
We can cancel \( \frac{3}{2} k \) from both sides:
\[ T_2 = \frac{1}{2} \times 300 \]

### Step 7: Calculate \( T_2 \)
Now, calculate \( T_2 \):
\[ T_2 = 150 \, K \]

### Step 8: Convert \( T_2 \) back to Celsius
To convert \( T_2 \) back to Celsius:
\[ T_2 = 150 - 273 = -123^\circ C \]

### Final Answer
The temperature at which the kinetic energy of gas molecules is half of the value at \( 27^\circ C \) is:
\[ -123^\circ C \]