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

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

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

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Critical Temperature For Clausius Model is the highest temperature at which the substance can exist as a liquid. At this phase boundaries vanish, the substance can exist both as a liquid and vapor. Check FAQs

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

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

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

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

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

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

215.5047 = \frac{(800 + (\frac{0.1}{({(22.4 + 0.0002)}^{2})})) \cdot (\frac{22.4 - 0.0024}{8.3145})}{10}

215.5047K = \frac{(800Pa + (\frac{0.1}{({(22.4m³/mol + 0.0002)}^{2})})) \cdot (\frac{22.4m³/mol - 0.0024}{8.3145})}{10}

215.5047 = \frac{(800 + (\frac{0.1}{({(22.4 + 0.0002)}^{2})})) \cdot (\frac{22.4 - 0.0024}{8.3145})}{10}

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

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

FIRST Step Consider the formula

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

Next Step Substitute values of Variables

∴

T' c = \frac{(800Pa + (\frac{0.1}{({(22.4m³/mol + 0.0002)}^{2})})) \cdot (\frac{22.4m³/mol - 0.0024}{[R]})}{10}

Next Step Substitute values of Constants

∴

T' c = \frac{(800Pa + (\frac{0.1}{({(22.4m³/mol + 0.0002)}^{2})})) \cdot (\frac{22.4m³/mol - 0.0024}{8.3145})}{10}

Next Step Prepare to Evaluate

∴

T' c = \frac{(800 + (\frac{0.1}{({(22.4 + 0.0002)}^{2})})) \cdot (\frac{22.4 - 0.0024}{8.3145})}{10}

Next Step Evaluate

∴

T' c = 215.504732976977K

LAST Step Rounding Answer

∴

T' c = 215.5047K

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

Variables

Constants

Critical Temperature For Clausius Model

Critical Temperature For Clausius Model is the highest temperature at which the substance can exist as a liquid. At this phase boundaries vanish, the substance can exist both as a liquid and vapor.

Symbol: T'_c

Measurement: TemperatureUnit: K

Note: Value should be greater than 0.

Formulas to find Critical Temperature For Clausius Model

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