At which of the following temperatures would the molecules of a gas have twice the average kinetic energy they have at `20^(@)C`?

Options

A. `40^(@)C`
B. `80^(@)C`
C. `313^(@)C`
D. `586^(@)C`

Answer

To find the temperature at which the molecules of a gas have twice the average kinetic energy they have at \(20^\circ C\), we can follow these steps:

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

### Step 2: Convert the given temperature from Celsius to Kelvin
The given temperature is \(20^\circ C\). To convert this to Kelvin:
\[ T = 20 + 273 = 293 \, K \]

### Step 3: Calculate the average kinetic energy at \(20^\circ C\)
Using the formula for average kinetic energy:
\[ E_1 = \frac{3}{2} k T_1 = \frac{3}{2} k (293) \]

### Step 4: Set up the equation for twice the average kinetic energy
We want to find the temperature \(T'\) at which the average kinetic energy is twice that at \(20^\circ C\):
\[ E_2 = 2E_1 \]
Substituting the expression for \(E_1\):
\[ E_2 = 2 \left( \frac{3}{2} k (293) \right) = 3 k (293) \]

### Step 5: Relate \(E_2\) to the new temperature \(T'\)
Using the formula for average kinetic energy again for the new temperature:
\[ E_2 = \frac{3}{2} k T' \]
Setting the two expressions for \(E_2\) equal to each other:
\[ 3 k (293) = \frac{3}{2} k T' \]

### Step 6: Solve for \(T'\)
We can cancel \(k\) from both sides and simplify:
\[ 3 (293) = \frac{3}{2} T' \]
Multiplying both sides by \(2\):
\[ 6 (293) = 3 T' \]
Dividing both sides by \(3\):
\[ T' = 2 (293) = 586 \, K \]

### Step 7: Convert the new temperature back to Celsius
To convert \(586 \, K\) back to Celsius:
\[ T' = 586 - 273 = 313 \, °C \]

### Conclusion
Thus, the temperature at which the molecules of a gas have twice the average kinetic energy they have at \(20^\circ C\) is \(313^\circ C\).