The temperature of a gas is `-68^(@)C`. To what temperature should it be heated so that the average kinetic energy of the molecules be doubled

Options

A. `137^@ C`
B. `127^@ C`
C. `100^@ C`
D. `105^@ C`

Answer

To solve the problem, we need to follow these steps:

### Step 1: Convert the initial temperature from Celsius to Kelvin
The initial temperature \( T_1 \) is given as \( -68^\circ C \). To convert this to Kelvin, we use the formula:
\[ T(K) = T(°C) + 273 \]
So,
\[ T_1 = -68 + 273 = 205 \, K \]

### Step 2: Understand the relationship between kinetic energy and temperature
The average kinetic energy of gas molecules is directly proportional to the absolute temperature (in Kelvin). This can be expressed as:
\[ KE \propto T \]
If the average kinetic energy is doubled, we can express this as:
\[ KE_2 = 2 \cdot KE_1 \]
This implies:
\[ \frac{KE_1}{KE_2} = \frac{T_1}{T_2} \]
Substituting \( KE_2 = 2 \cdot KE_1 \):
\[ \frac{KE_1}{2 \cdot KE_1} = \frac{T_1}{T_2} \]
This simplifies to:
\[ \frac{1}{2} = \frac{T_1}{T_2} \]

### Step 3: Solve for the final temperature \( T_2 \)
Rearranging the equation gives:
\[ T_2 = 2 \cdot T_1 \]
Substituting the value of \( T_1 \):
\[ T_2 = 2 \cdot 205 \, K = 410 \, K \]

### Step 4: Convert the final temperature back to Celsius
Now, we convert \( T_2 \) back to Celsius:
\[ T(°C) = T(K) - 273 \]
So,
\[ T_2(°C) = 410 - 273 = 137^\circ C \]

### Final Answer
The temperature to which the gas should be heated so that the average kinetic energy of the molecules is doubled is \( 137^\circ C \).

---