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Average Thermal Energy of Non-linear Polyatomic Gas Molecule Formula

GoAverage Thermal Energy of Linear Polyatomic Gas Molecule Formula

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Thermal Energy is the input heat energy to a given system. This input heat energy is converted into useful work and a part of it is wasted in doing so. Check FAQs

Q in = ((\frac{3}{2}) \cdot [BoltZ] \cdot T) + ((0.5 \cdot I y \cdot ({ω y}^{2})) + (0.5 \cdot I z \cdot ({ω z}^{2}))) + ((3 \cdot N) - 6) \cdot ([BoltZ] \cdot T)

Q_in - Thermal Energy?T - Temperature?I_y - Moment of Inertia along Y-axis?ω_y - Angular Velocity along Y-axis?I_z - Moment of Inertia along Z-axis?ω_z - Angular Velocity along Z-axis?N - Atomicity?[BoltZ] - Boltzmann constant?[BoltZ] - Boltzmann constant?

Average Thermal Energy of Non-linear Polyatomic Gas Molecule Example

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Here is how the Average Thermal Energy of Non-linear Polyatomic Gas Molecule equation looks like with Values.

Here is how the Average Thermal Energy of Non-linear Polyatomic Gas Molecule equation looks like with Units.

Here is how the Average Thermal Energy of Non-linear Polyatomic Gas Molecule equation looks like.

27.0348 = ((\frac{3}{2}) \cdot 1.4E-23 \cdot 85) + ((0.5 \cdot 60 \cdot (35^{2})) + (0.5 \cdot 65 \cdot (40^{2}))) + ((3 \cdot 3) - 6) \cdot (1.4E-23 \cdot 85)

27.0348J = ((\frac{3}{2}) \cdot 1.4E-23J/K \cdot 85K) + ((0.5 \cdot 60kg·m² \cdot ({35degree/s}^{2})) + (0.5 \cdot 65kg·m² \cdot ({40degree/s}^{2}))) + ((3 \cdot 3) - 6) \cdot (1.4E-23J/K \cdot 85K)

27.0348 = ((\frac{3}{2}) \cdot 1.4E-23 \cdot 85) + ((0.5 \cdot 60 \cdot (35^{2})) + (0.5 \cdot 65 \cdot (40^{2}))) + ((3 \cdot 3) - 6) \cdot (1.4E-23 \cdot 85)

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Home » Chemistry » Kinetic Theory of Gases » Equipartition Principle and Heat Capacity » Average Thermal Energy of Non-linear Polyatomic Gas Molecule

Average Thermal Energy of Non-linear Polyatomic Gas Molecule Solution

Follow our step by step solution on how to calculate Average Thermal Energy of Non-linear Polyatomic Gas Molecule?

FIRST Step Consider the formula

Q in = ((\frac{3}{2}) \cdot [BoltZ] \cdot T) + ((0.5 \cdot I y \cdot ({ω y}^{2})) + (0.5 \cdot I z \cdot ({ω z}^{2}))) + ((3 \cdot N) - 6) \cdot ([BoltZ] \cdot T)

Next Step Substitute values of Variables

∴

Q in = ((\frac{3}{2}) \cdot [BoltZ] \cdot 85K) + ((0.5 \cdot 60kg·m² \cdot ({35degree/s}^{2})) + (0.5 \cdot 65kg·m² \cdot ({40degree/s}^{2}))) + ((3 \cdot 3) - 6) \cdot ([BoltZ] \cdot 85K)

Next Step Substitute values of Constants

∴

Q in = ((\frac{3}{2}) \cdot 1.4E-23J/K \cdot 85K) + ((0.5 \cdot 60kg·m² \cdot ({35degree/s}^{2})) + (0.5 \cdot 65kg·m² \cdot ({40degree/s}^{2}))) + ((3 \cdot 3) - 6) \cdot (1.4E-23J/K \cdot 85K)

Next Step Convert Units

∴

Q in = ((\frac{3}{2}) \cdot 1.4E-23J/K \cdot 85K) + ((0.5 \cdot 60kg·m² \cdot ({0.6109rad/s}^{2})) + (0.5 \cdot 65kg·m² \cdot ({0.6981rad/s}^{2}))) + ((3 \cdot 3) - 6) \cdot (1.4E-23J/K \cdot 85K)

Next Step Prepare to Evaluate

∴

Q in = ((\frac{3}{2}) \cdot 1.4E-23 \cdot 85) + ((0.5 \cdot 60 \cdot ({0.6109}^{2})) + (0.5 \cdot 65 \cdot ({0.6981}^{2}))) + ((3 \cdot 3) - 6) \cdot (1.4E-23 \cdot 85)

Next Step Evaluate

∴

Q in = 27.0347960060603J

LAST Step Rounding Answer

∴

Q in = 27.0348J

Average Thermal Energy of Non-linear Polyatomic Gas Molecule Formula Elements

Variables

Constants

Thermal Energy

Thermal Energy is the input heat energy to a given system. This input heat energy is converted into useful work and a part of it is wasted in doing so.

Symbol: Q_in

Measurement: EnergyUnit: J

Note: Value should be greater than 0.

Temperature

Temperature is the degree or intensity of heat present in a substance or object.

Symbol: T

Measurement: TemperatureUnit: K