Temperature of an ideal 'gas whose molecules have average kinetic energy 1 e V is

Options

A. 3590 K
B. 7730 K
C. 4460 K
D. 5197 K

Answer

To find the temperature of an ideal gas whose molecules have an average kinetic energy of 1 eV, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the relationship between kinetic energy and temperature**:
For a monoatomic ideal gas, the average kinetic energy (KE) per molecule is given by the formula:
\[ KE = \frac{3}{2} k T \]
where \( k \) is the Boltzmann constant and \( T \) is the temperature in Kelvin.

2. **Convert the given kinetic energy from eV to Joules**:
The average kinetic energy is given as 1 eV. To convert electron volts to Joules, we use the conversion factor:
\[ 1 \text{ eV} = 1.602 \times 10^{-19} \text{ J} \]
Therefore,
\[ KE = 1 \text{ eV} = 1.602 \times 10^{-19} \text{ J} \]

3. **Rearrange the kinetic energy formula to solve for temperature**:
From the kinetic energy formula, we can express temperature \( T \) as:
\[ T = \frac{2 \cdot KE}{3k} \]

4. **Substitute the values into the equation**:
We know:
- \( KE = 1.602 \times 10^{-19} \text{ J} \)
- The Boltzmann constant \( k = 1.38 \times 10^{-23} \text{ J/K} \)

Plugging these values into the equation for temperature:
\[ T = \frac{2 \cdot (1.602 \times 10^{-19})}{3 \cdot (1.38 \times 10^{-23})} \]

5. **Calculate the numerator and denominator**:
- Numerator:
\[ 2 \cdot (1.602 \times 10^{-19}) = 3.204 \times 10^{-19} \]
- Denominator:
\[ 3 \cdot (1.38 \times 10^{-23}) = 4.14 \times 10^{-23} \]

6. **Divide the numerator by the denominator**:
\[ T = \frac{3.204 \times 10^{-19}}{4.14 \times 10^{-23}} \approx 7730.77 \text{ K} \]

7. **Final result**:
The temperature of the ideal gas is approximately \( 7730 \text{ K} \).