Question

The average kinetic energy of a monatomic molecule is 0.414 eV at temperature: (Use \( k_B = 1.38 \times 10^{-23} \, \text{J/mol-K} \))

• A. 3000 K
• B. 3200 K
• C. 1600 K
• D. 1500 K

Answer 1

The Correct Option is B

To find the temperature at which the average kinetic energy of a monatomic molecule is 0.414 eV, we can use the formula that relates the average kinetic energy of a gas molecule to the temperature. The average kinetic energy per molecule for a monatomic ideal gas is given by the equation:

\(KE_{\text{avg}} = \frac{3}{2} k_B T\) 

Where:

• \(KE_{\text{avg}}\) is the average kinetic energy per molecule.
• \(k_B\) is the Boltzmann constant, given as \(1.38 \times 10^{-23} \, \text{J/K}\).
• \(T\) is the absolute temperature in Kelvin.

First, convert the given kinetic energy from electronvolts (eV) to joules (J):

\(1 \, \text{eV} = 1.602 \times 10^{-19} \, \text{J}\)

Therefore,

\(0.414 \, \text{eV} = 0.414 \times 1.602 \times 10^{-19} \, \text{J} = 6.633 \times 10^{-20} \, \text{J}\)

Now, substitute this value into the equation for average kinetic energy:

\(\frac{3}{2} k_B T = 6.633 \times 10^{-20}\)

Solving for \(T\):

\(T = \frac{2 \times 6.633 \times 10^{-20}}{3 \times 1.38 \times 10^{-23}}\)

\(T = \frac{1.3266 \times 10^{-19}}{4.14 \times 10^{-23}}\)

\(T = 3204.35 \, \text{K}\)

Rounded to the nearest whole number, this gives \(T = 3200 \, \text{K}\).

Thus, the correct answer is 3200 K.

Answer 2

For a monatomic gas, the average kinetic energy per molecule is:

\[ KE = \frac{3}{2} k_B T \]

Given \( KE = 0.414 \, \text{eV} \), convert this to joules:

\[ 0.414 \, \text{eV} = 0.414 \times 1.6 \times 10^{-19} \, \text{J} = 6.624 \times 10^{-20} \, \text{J} \]

Now,

\[ 6.624 \times 10^{-20} = \frac{3}{2} \times 1.38 \times 10^{-23} \times T \]

Solving for \( T \):

\[ T = \frac{6.624 \times 10^{-20}}{\left(\frac{3}{2}\right) \times 1.38 \times 10^{-23}} \approx 3200 \, \text{K} \]