Calculate the value of kT for 300 K, 0°C, and 373 K. Express the answers in both eV and J
The Correct Answer and Explanation is :
To calculate the value of (kT) (the thermal energy) at different temperatures, we use the formula:
[
kT = k \times T
]
Where:
- (k) is the Boltzmann constant, which is approximately (1.38 \times 10^{-23}) J/K or (8.617 \times 10^{-5}) eV/K.
- (T) is the temperature in Kelvin (K).
We’ll calculate (kT) for the three given temperatures: 300 K, 0°C, and 373 K.
1. For T = 300 K:
Using the value of (k = 1.38 \times 10^{-23}) J/K:
[
kT = (1.38 \times 10^{-23} \, \text{J/K}) \times (300 \, \text{K}) = 4.14 \times 10^{-21} \, \text{J}
]
To convert to electron volts (eV), we use (1 \, \text{eV} = 1.602 \times 10^{-19} \, \text{J}):
[
kT = \frac{4.14 \times 10^{-21} \, \text{J}}{1.602 \times 10^{-19} \, \text{J/eV}} = 0.0258 \, \text{eV}
]
2. For T = 0°C (273.15 K):
We first convert 0°C to Kelvin: (T = 273.15 \, \text{K}).
[
kT = (1.38 \times 10^{-23} \, \text{J/K}) \times (273.15 \, \text{K}) = 3.77 \times 10^{-21} \, \text{J}
]
Convert to eV:
[
kT = \frac{3.77 \times 10^{-21} \, \text{J}}{1.602 \times 10^{-19} \, \text{J/eV}} = 0.0235 \, \text{eV}
]
3. For T = 373 K:
[
kT = (1.38 \times 10^{-23} \, \text{J/K}) \times (373 \, \text{K}) = 5.15 \times 10^{-21} \, \text{J}
]
Convert to eV:
[
kT = \frac{5.15 \times 10^{-21} \, \text{J}}{1.602 \times 10^{-19} \, \text{J/eV}} = 0.0322 \, \text{eV}
]
Summary of Results:
- At 300 K:
( kT = 4.14 \times 10^{-21} \, \text{J} ) or ( 0.0258 \, \text{eV} ) - At 0°C (273.15 K):
( kT = 3.77 \times 10^{-21} \, \text{J} ) or ( 0.0235 \, \text{eV} ) - At 373 K:
( kT = 5.15 \times 10^{-21} \, \text{J} ) or ( 0.0322 \, \text{eV} )
Explanation:
The quantity (kT) represents the average thermal energy per particle in a system at a given temperature. The Boltzmann constant, (k), links temperature to energy, and it expresses how much energy corresponds to a temperature of one kelvin. At room temperature (about 300 K), the thermal energy is on the order of 0.025 eV. The value of (kT) increases with temperature, which is why at higher temperatures like 373 K, the thermal energy is greater.
In these calculations, we used the SI unit for energy, the Joule, and the electron volt (eV), a common unit in physics, particularly in particle physics, because it is convenient when dealing with energies at the atomic or molecular scale.