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Average Thermal Energy of Linear Polyatomic Gas Molecule Formula

GoAverage Thermal Energy of Non-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) - 5) \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 Linear Polyatomic Gas Molecule Example

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

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

Here is how the Average Thermal Energy of 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) - 5) \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) - 5) \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) - 5) \cdot (1.4E-23 \cdot 85)

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Average Thermal Energy of Linear Polyatomic Gas Molecule Solution

Follow our step by step solution on how to calculate Average Thermal Energy of 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) - 5) \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) - 5) \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) - 5) \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) - 5) \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) - 5) \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 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.

Formulas to find Thermal Energy

Temperature

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

Symbol: T

Measurement: TemperatureUnit: K