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Reduced Temperature of Real Gas using Clausius Equation given Reduced and Critical 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 r \cdot P' c) + (\frac{a}{({((V' m,r \cdot V m,c) + c)}^{2})})) \cdot (\frac{(V' m,r \cdot V m,c) - b'}{[R]})}{T' c}

T_r - Reduced Temperature?P_r - Reduced Pressure?P'_c - Critical Pressure of Real Gas?a - Clausius Parameter a?V'_m,r - Reduced Molar Volume for Real Gas?V_m,c - Critical 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 Reduced and Critical Parameters Example

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

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

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

295366.982 = \frac{((0.8 \cdot 4.6E+6) + (\frac{0.1}{({((8.96 \cdot 11.5) + 0.0002)}^{2})})) \cdot (\frac{(8.96 \cdot 11.5) - 0.0024}{8.3145})}{154.4}

295366.982 = \frac{((0.8 \cdot 4.6E+6Pa) + (\frac{0.1}{({((8.96 \cdot 11.5m³/mol) + 0.0002)}^{2})})) \cdot (\frac{(8.96 \cdot 11.5m³/mol) - 0.0024}{8.3145})}{154.4K}

295366.982 = \frac{((0.8 \cdot 4.6E+6) + (\frac{0.1}{({((8.96 \cdot 11.5) + 0.0002)}^{2})})) \cdot (\frac{(8.96 \cdot 11.5) - 0.0024}{8.3145})}{154.4}

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

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

FIRST Step Consider the formula

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

Next Step Substitute values of Variables

∴

T r = \frac{((0.8 \cdot 4.6E+6Pa) + (\frac{0.1}{({((8.96 \cdot 11.5m³/mol) + 0.0002)}^{2})})) \cdot (\frac{(8.96 \cdot 11.5m³/mol) - 0.0024}{[R]})}{154.4K}

Next Step Substitute values of Constants

∴

T r = \frac{((0.8 \cdot 4.6E+6Pa) + (\frac{0.1}{({((8.96 \cdot 11.5m³/mol) + 0.0002)}^{2})})) \cdot (\frac{(8.96 \cdot 11.5m³/mol) - 0.0024}{8.3145})}{154.4K}

Next Step Prepare to Evaluate

∴

T r = \frac{((0.8 \cdot 4.6E+6) + (\frac{0.1}{({((8.96 \cdot 11.5) + 0.0002)}^{2})})) \cdot (\frac{(8.96 \cdot 11.5) - 0.0024}{8.3145})}{154.4}

Next Step Evaluate

∴

T r = 295366.982012086

LAST Step Rounding Answer

∴

T r = 295366.982

Reduced Temperature of Real Gas using Clausius Equation given Reduced and Critical 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

Reduced Pressure

Reduced Pressure is the ratio of the actual pressure of the fluid to its critical pressure. It is dimensionless.

Symbol: P_r

Measurement: NAUnit: Unitless

Note: Value should be between 0 to 1.

Formulas to find Reduced Pressure

Critical Pressure of Real Gas

Critical Pressure of Real Gas is the minimum pressure required to liquify a substance at the critical temperature.

Symbol: P'_c

Measurement: PressureUnit: Pa

Note: Value should be greater than 0.

Formulas to find Critical Pressure of Real Gas

Clausius Parameter a

Clausius parameter a is an empirical parameter characteristic to equation obtained from Clausius model of real gas.

Symbol: a

Measurement: NAUnit: Unitless