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

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

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

T_{r_RP_AP} - Reduced Temperature given RP AP?p - Pressure?a - Clausius Parameter a?V_m - Molar Volume?c - Clausius Parameter c?b' - Clausius Parameter b for Real Gas?T_rg - Temperature of Real Gas?[R] - Universal gas constant?

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

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

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

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

7.1835 = \frac{(800 + (\frac{0.1}{({(22.4 + 0.0002)}^{2})})) \cdot (\frac{22.4 - 0.0024}{8.3145})}{300}

7.1835 = \frac{(800Pa + (\frac{0.1}{({(22.4m³/mol + 0.0002)}^{2})})) \cdot (\frac{22.4m³/mol - 0.0024}{8.3145})}{300K}

7.1835 = \frac{(800 + (\frac{0.1}{({(22.4 + 0.0002)}^{2})})) \cdot (\frac{22.4 - 0.0024}{8.3145})}{300}

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

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

FIRST Step Consider the formula

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

Next Step Substitute values of Variables

∴

T r_RP_AP = \frac{(800Pa + (\frac{0.1}{({(22.4m³/mol + 0.0002)}^{2})})) \cdot (\frac{22.4m³/mol - 0.0024}{[R]})}{300K}

Next Step Substitute values of Constants

∴

T r_RP_AP = \frac{(800Pa + (\frac{0.1}{({(22.4m³/mol + 0.0002)}^{2})})) \cdot (\frac{22.4m³/mol - 0.0024}{8.3145})}{300K}

Next Step Prepare to Evaluate

∴

T r_RP_AP = \frac{(800 + (\frac{0.1}{({(22.4 + 0.0002)}^{2})})) \cdot (\frac{22.4 - 0.0024}{8.3145})}{300}

Next Step Evaluate

∴

T r_RP_AP = 7.18349109923257

LAST Step Rounding Answer

∴

T r_RP_AP = 7.1835

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

Variables

Constants

Reduced Temperature given RP AP

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

Symbol: T_{r_RP_AP}

Measurement: NAUnit: Unitless

Note: Value can be positive or negative.

Formulas to find Reduced Temperature given RP AP

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

Note: Value can be positive or negative.

Formulas to find Pressure

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

Note: Value can be positive or negative.

Formulas to find Clausius Parameter a

Molar Volume

Molar Volume is the volume occupied by one mole of a real gas at standard temperature and pressure.

Symbol: V_m

Measurement: Molar Magnetic SusceptibilityUnit: m³/mol

Note: Value can be positive or negative.

Formulas to find Molar Volume

Clausius Parameter c

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

Symbol: c

Measurement: NAUnit: Unitless

Note: Value can be positive or negative.

Formulas to find Clausius Parameter c

Clausius Parameter b for Real Gas

Clausius Parameter b for Real Gas is an empirical parameter characteristic to equation obtained from Clausius model of real gas.

Symbol: b'

Measurement: NAUnit: Unitless

Note: Value can be positive or negative.

Formulas to find Clausius Parameter b for Real Gas

Temperature of Real Gas

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

Symbol: T_rg

Measurement: TemperatureUnit: K

Note: Value can be positive or negative.

Formulas to find Temperature of Real Gas

Universal gas constant

Universal gas constant is a fundamental physical constant that appears in the ideal gas law, relating the pressure, volume, and temperature of an ideal gas.

Symbol: [R]

Value: 8.31446261815324

Other formulas in Reduced Temperature of Real Gas category

Go Actual Pressure of Real Gas given Clausius Parameter a, Reduced and Critical Parameters

Pa = (\frac{27 \cdot ({[R]}^{2}) \cdot ({T' c}^{3})}{64 \cdot a}) \cdot P r

Go Actual Pressure of Real Gas given Clausius Parameter b, Reduced and Actual Parameters

Pb = (\frac{[R] \cdot (\frac{T rg}{T r})}{4 \cdot ((\frac{V real}{V r}) - b')}) \cdot P r

How to Evaluate Reduced Temperature of Real Gas using Clausius Equation given Reduced and Actual Parameters?

Reduced Temperature of Real Gas using Clausius Equation given Reduced and Actual Parameters evaluator uses Reduced Temperature given RP AP = ((Pressure+(Clausius Parameter a/(((Molar Volume+Clausius Parameter c)^2))))*((Molar Volume-Clausius Parameter b for Real Gas)/[R]))/Temperature of Real Gas to evaluate the Reduced Temperature given RP AP, The Reduced Temperature of real gas using Clausius equation given reduced and actual parameters formula is defined as the ratio of the actual temperature of the fluid to its critical temperature. It is dimensionless. Reduced Temperature given RP AP is denoted by T_{r_RP_AP} symbol.

How to evaluate Reduced Temperature of Real Gas using Clausius Equation given Reduced and Actual Parameters using this online evaluator? To use this online evaluator for Reduced Temperature of Real Gas using Clausius Equation given Reduced and Actual Parameters, enter Pressure (p), Clausius Parameter a (a), Molar Volume (V_m), Clausius Parameter c (c), Clausius Parameter b for Real Gas (b') & Temperature of Real Gas (T_rg) and hit the calculate button.

FAQs on Reduced Temperature of Real Gas using Clausius Equation given Reduced and Actual Parameters

What is the formula to find Reduced Temperature of Real Gas using Clausius Equation given Reduced and Actual Parameters?

The formula of Reduced Temperature of Real Gas using Clausius Equation given Reduced and Actual Parameters is expressed as Reduced Temperature given RP AP = ((Pressure+(Clausius Parameter a/(((Molar Volume+Clausius Parameter c)^2))))*((Molar Volume-Clausius Parameter b for Real Gas)/[R]))/Temperature of Real Gas. Here is an example- 25.3535 = ((800+(0.1/(((22.4+0.0002)^2))))*((22.4-0.00243)/[R]))/300.

How to calculate Reduced Temperature of Real Gas using Clausius Equation given Reduced and Actual Parameters?

With Pressure (p), Clausius Parameter a (a), Molar Volume (V_m), Clausius Parameter c (c), Clausius Parameter b for Real Gas (b') & Temperature of Real Gas (T_rg) we can find Reduced Temperature of Real Gas using Clausius Equation given Reduced and Actual Parameters using the formula - Reduced Temperature given RP AP = ((Pressure+(Clausius Parameter a/(((Molar Volume+Clausius Parameter c)^2))))*((Molar Volume-Clausius Parameter b for Real Gas)/[R]))/Temperature of Real Gas. This formula also uses Universal gas constant .

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

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

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

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

Other formulas in Reduced Temperature of Real Gas category

How to Evaluate Reduced Temperature of Real Gas using Clausius Equation given Reduced and Actual Parameters?

FAQs on Reduced Temperature of Real Gas using Clausius Equation given Reduced and Actual Parameters

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