Question 260: Thermodynamics & KTG

Question

The average kinetic energy of a monatomic molecule is $$0.414 \mathrm{~eV}$$ at temperature :
(Use $$K_B=1.38 	imes 10^{-23} \mathrm{~J} / \mathrm{mol}-\mathrm{K}$$)

Accepted Answer

To find the temperature at which the average kinetic energy of a monatomic molecule is $$0.414 \, 	ext{eV}$$, we use the equation for the average kinetic energy of a molecule in terms of temperature:
$$K_{	ext{avg}} = \frac{3}{2} k_B T$$
Where:
$$K_{	ext{avg}}$$ is the average kinetic energy
$$k_B$$ is the Boltzmann constant, given as $$1.38 	imes 10^{-23} \, 	ext{J/K}$$
$$T$$ is the temperature in Kelvin
First, we need to convert the average kinetic energy from eV to Joules since the Boltzmann constant is in Joules. The conversion factor is $$1 \, 	ext{eV} = 1.6 	imes 10^{-19} \, 	ext{J}$$, so:
$$K_{	ext{avg}} = 0.414 \, 	ext{eV} = 0.414 	imes 1.6 	imes 10^{-19} \, 	ext{J}$$
Substituting $$K_{	ext{avg}}$$ and $$k_B$$ into the equation:
$$\frac{3}{2} k_B T = 0.414 	imes 1.6 	imes 10^{-19} \, 	ext{J}$$
Solving for $$T$$:
$$T = \frac{0.414 	imes 1.6 	imes 10^{-19} \, 	ext{J}}{\frac{3}{2} 	imes 1.38 	imes 10^{-23} \, 	ext{J/K}}$$
After performing the calculations:
$$T \approx 3200 \, 	ext{K}$$
Therefore, the temperature at which the average kinetic energy of a monatomic molecule is $$0.414 \, 	ext{eV}$$ is approximately 3200 K.

Answer

3000 K

Answer

3200 K

Answer

1600 K

Answer

1500 K

Thermodynamics & KTG

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

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