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Reduced Temperature of Real Gas using Clausius Equation given Critical and Actual Parameters Formula

GoReduced Temperature of Real Gas given Clausius Parameter and Actual Parameters Formula

GoReduced Temperature of Real Gas given Clausius Parameter a, Reduced and Actual Parameters Formula

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Reduced Temperature is the ratio of the actual temperature of the fluid to its critical temperature. It is dimensionless. Check FAQs

T r = \frac{(p + (\frac{a}{({(V m + c)}^{2})})) \cdot (\frac{V m - b'}{[R]})}{T' c}

T_r - Reduced Temperature?p - Pressure?a - Clausius Parameter a?V_m - Molar Volume?c - Clausius Parameter c?b' - Clausius Parameter b for Real Gas?T'_c - Critical Temperature For Clausius Model?[R] - Universal gas constant?

Reduced Temperature of Real Gas using Clausius Equation given Critical and Actual Parameters Example

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Here is how the Reduced Temperature of Real Gas using Clausius Equation given Critical and Actual Parameters equation looks like with Values.

Here is how the Reduced Temperature of Real Gas using Clausius Equation given Critical and Actual Parameters equation looks like with Units.

Here is how the Reduced Temperature of Real Gas using Clausius Equation given Critical and Actual Parameters equation looks like.

13.9576 = \frac{(800 + (\frac{0.1}{({(22.4 + 0.0002)}^{2})})) \cdot (\frac{22.4 - 0.0024}{8.3145})}{154.4}

13.9576 = \frac{(800Pa + (\frac{0.1}{({(22.4m³/mol + 0.0002)}^{2})})) \cdot (\frac{22.4m³/mol - 0.0024}{8.3145})}{154.4K}

13.9576 = \frac{(800 + (\frac{0.1}{({(22.4 + 0.0002)}^{2})})) \cdot (\frac{22.4 - 0.0024}{8.3145})}{154.4}

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Reduced Temperature of Real Gas using Clausius Equation given Critical and Actual Parameters Solution

Follow our step by step solution on how to calculate Reduced Temperature of Real Gas using Clausius Equation given Critical and Actual Parameters?

FIRST Step Consider the formula

T r = \frac{(p + (\frac{a}{({(V m + c)}^{2})})) \cdot (\frac{V m - b'}{[R]})}{T' c}

Next Step Substitute values of Variables

∴

T r = \frac{(800Pa + (\frac{0.1}{({(22.4m³/mol + 0.0002)}^{2})})) \cdot (\frac{22.4m³/mol - 0.0024}{[R]})}{154.4K}

Next Step Substitute values of Constants

∴

T r = \frac{(800Pa + (\frac{0.1}{({(22.4m³/mol + 0.0002)}^{2})})) \cdot (\frac{22.4m³/mol - 0.0024}{8.3145})}{154.4K}

Next Step Prepare to Evaluate

∴

T r = \frac{(800 + (\frac{0.1}{({(22.4 + 0.0002)}^{2})})) \cdot (\frac{22.4 - 0.0024}{8.3145})}{154.4}

Next Step Evaluate

∴

T r = 13.95756042597

LAST Step Rounding Answer

∴

T r = 13.9576

Reduced Temperature of Real Gas using Clausius Equation given Critical and Actual Parameters Formula Elements

Variables

Constants

Reduced Temperature

Reduced Temperature is the ratio of the actual temperature of the fluid to its critical temperature. It is dimensionless.

Symbol: T_r

Measurement: NAUnit: Unitless

Note: Value should be greater than 0.

Formulas to find Reduced Temperature

Pressure

Pressure is the force applied perpendicular to the surface of an object per unit area over which that force is distributed.

Symbol: p

Measurement: PressureUnit: Pa