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`

Explanation

To find the temperature at which the average kinetic energy of gas molecules is doubled compared to their average kinetic energy at \(20^{\circ}C\), we can follow these steps:

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

### Step 2: Convert \(20^{\circ}C\) to Kelvin
To convert Celsius to Kelvin, we use the formula:
\[ T(K) = T(°C) + 273.15 \]
For \(20^{\circ}C\):
\[ T = 20 + 273.15 = 293.15 \, K \]

### Step 3: Calculate the average kinetic energy at \(20^{\circ}C\)
Using the temperature we just calculated:
\[ KE_{20°C} = \frac{3}{2} k_B (293.15) \]

### Step 4: Determine the temperature for doubled kinetic energy
If we want the average kinetic energy to be doubled, we set up the equation:
\[ KE_{new} = 2 \times KE_{20°C} \]
Substituting the expression for kinetic energy:
\[ \frac{3}{2} k_B T_{new} = 2 \left(\frac{3}{2} k_B (293.15)\right) \]

### Step 5: Simplify the equation
We can cancel \(\frac{3}{2} k_B\) from both sides:
\[ T_{new} = 2 \times 293.15 \]

### Step 6: Calculate \(T_{new}\)
Now, calculate \(T_{new}\):
\[ T_{new} = 586.3 \, K \]

### Step 7: Convert \(T_{new}\) back to Celsius (if needed)
To convert back to Celsius:
\[ T(°C) = T(K) - 273.15 \]
Thus,
\[ T_{new}(°C) = 586.3 - 273.15 \approx 313.15 \, °C \]

### Final Answer
The temperature at which the molecules of a gas would have twice the average kinetic energy they have at \(20^{\circ}C\) is approximately \(586.3 \, K\) or \(313.15 \, °C\).

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