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Enthalpy using Integrated Form of Clausius-Clapeyron Equation Formula

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Change in enthalpy is the thermodynamic quantity equivalent to the total difference between the heat content of a system. Check FAQs

ΔH = \frac{- \ln (\frac{P f}{P i}) \cdot [R]}{(\frac{1}{T f}) - (\frac{1}{T i})}

ΔH - Change in Enthalpy?P_f - Final Pressure of System?P_i - Initial Pressure of System?T_f - Final Temperature?T_i - Initial Temperature?[R] - Universal gas constant?

Enthalpy using Integrated Form of Clausius-Clapeyron Equation Example

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Here is how the Enthalpy using Integrated Form of Clausius-Clapeyron Equation equation looks like with Values.

Here is how the Enthalpy using Integrated Form of Clausius-Clapeyron Equation equation looks like with Units.

Here is how the Enthalpy using Integrated Form of Clausius-Clapeyron Equation equation looks like.

25020.2946 = \frac{- \ln (\frac{133.07}{65}) \cdot 8.3145}{(\frac{1}{700}) - (\frac{1}{600})}

25020.2946J/kg = \frac{- \ln (\frac{133.07Pa}{65Pa}) \cdot 8.3145}{(\frac{1}{700K}) - (\frac{1}{600K})}

25020.2946 = \frac{- \ln (\frac{133.07}{65}) \cdot 8.3145}{(\frac{1}{700}) - (\frac{1}{600})}

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Enthalpy using Integrated Form of Clausius-Clapeyron Equation Solution

Follow our step by step solution on how to calculate Enthalpy using Integrated Form of Clausius-Clapeyron Equation?

FIRST Step Consider the formula

ΔH = \frac{- \ln (\frac{P f}{P i}) \cdot [R]}{(\frac{1}{T f}) - (\frac{1}{T i})}

Next Step Substitute values of Variables

∴

ΔH = \frac{- \ln (\frac{133.07Pa}{65Pa}) \cdot [R]}{(\frac{1}{700K}) - (\frac{1}{600K})}

Next Step Substitute values of Constants

∴

ΔH = \frac{- \ln (\frac{133.07Pa}{65Pa}) \cdot 8.3145}{(\frac{1}{700K}) - (\frac{1}{600K})}

Next Step Prepare to Evaluate

∴

ΔH = \frac{- \ln (\frac{133.07}{65}) \cdot 8.3145}{(\frac{1}{700}) - (\frac{1}{600})}

Next Step Evaluate

∴

ΔH = 25020.2945531668J/kg

LAST Step Rounding Answer

∴

ΔH = 25020.2946J/kg

Enthalpy using Integrated Form of Clausius-Clapeyron Equation Formula Elements

Variables

Constants

Functions

Change in Enthalpy

Change in enthalpy is the thermodynamic quantity equivalent to the total difference between the heat content of a system.

Symbol: ΔH

Measurement: Heat of Combustion (per Mass)Unit: J/kg

Note: Value can be positive or negative.

Formulas to find Change in Enthalpy

Final Pressure of System

Final Pressure of System is the total final pressure exerted by the molecules inside the system.

Symbol: P_f

Measurement: PressureUnit: Pa

Note: Value can be positive or negative.

Formulas to find Final Pressure of System

Initial Pressure of System

Initial Pressure of System is the total initial pressure exerted by the molecules inside the system.

Symbol: P_i

Measurement: PressureUnit: Pa

Note: Value can be positive or negative.

Formulas to find Initial Pressure of System

Final Temperature

The Final temperature is the temperature at which measurements are made in final state.

Symbol: T_f

Measurement: TemperatureUnit: K

Note: Value should be greater than 0.

Formulas to find Final Temperature

Initial Temperature

The Initial temperature is defined as the measure of heat under initial state or conditions.

Symbol: T_i

Measurement: TemperatureUnit: K

Note: Value should be greater than 0.

Formulas to find Initial Temperature

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

The natural logarithm, also known as the logarithm to the base e, is the inverse function of the natural exponential function.

Syntax: ln(Number)

Other formulas in Clausius Clapeyron Equation category

Go August Roche Magnus Formula

e s = 6.1094 \cdot \exp (\frac{17.625 \cdot T}{T + 243.04})

Go Boiling Point given Enthalpy using Trouton's Rule

bp = \frac{H}{10.5 \cdot [R]}

Go Boiling Point using Trouton's Rule given Latent Heat

bp = \frac{LH}{10.5 \cdot [R]}

Go Boiling Point using Trouton's Rule given Specific Latent Heat

bp = \frac{L \cdot MW}{10.5 \cdot [R]}

How to Evaluate Enthalpy using Integrated Form of Clausius-Clapeyron Equation?

Enthalpy using Integrated Form of Clausius-Clapeyron Equation evaluator uses Change in Enthalpy = (-ln(Final Pressure of System/Initial Pressure of System)*[R])/((1/Final Temperature)-(1/Initial Temperature)) to evaluate the Change in Enthalpy, The Enthalpy using integrated form of Clausius-Clapeyron Equation is the difference in heat on the final and initial state of the system. Change in Enthalpy is denoted by ΔH symbol.

How to evaluate Enthalpy using Integrated Form of Clausius-Clapeyron Equation using this online evaluator? To use this online evaluator for Enthalpy using Integrated Form of Clausius-Clapeyron Equation, enter Final Pressure of System (P_f), Initial Pressure of System (P_i), Final Temperature (T_f) & Initial Temperature (T_i) and hit the calculate button.

FAQs on Enthalpy using Integrated Form of Clausius-Clapeyron Equation

What is the formula to find Enthalpy using Integrated Form of Clausius-Clapeyron Equation?

The formula of Enthalpy using Integrated Form of Clausius-Clapeyron Equation is expressed as Change in Enthalpy = (-ln(Final Pressure of System/Initial Pressure of System)*[R])/((1/Final Temperature)-(1/Initial Temperature)). Here is an example- -44014.366316 = (-ln(133.07/65)*[R])/((1/700)-(1/600)).

How to calculate Enthalpy using Integrated Form of Clausius-Clapeyron Equation?

With Final Pressure of System (P_f), Initial Pressure of System (P_i), Final Temperature (T_f) & Initial Temperature (T_i) we can find Enthalpy using Integrated Form of Clausius-Clapeyron Equation using the formula - Change in Enthalpy = (-ln(Final Pressure of System/Initial Pressure of System)*[R])/((1/Final Temperature)-(1/Initial Temperature)). This formula also uses Universal gas constant and Natural Logarithm (ln) function(s).

Can the Enthalpy using Integrated Form of Clausius-Clapeyron Equation be negative?

Yes, the Enthalpy using Integrated Form of Clausius-Clapeyron Equation, measured in Heat of Combustion (per Mass) can be negative.

Which unit is used to measure Enthalpy using Integrated Form of Clausius-Clapeyron Equation?

Enthalpy using Integrated Form of Clausius-Clapeyron Equation is usually measured using the Joule per Kilogram[J/kg] for Heat of Combustion (per Mass). Kilojoule per Kilogram[J/kg], Calorie (IT) per Gram[J/kg] are the few other units in which Enthalpy using Integrated Form of Clausius-Clapeyron Equation can be measured.

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Enthalpy using Integrated Form of Clausius-Clapeyron Equation Formula

Enthalpy using Integrated Form of Clausius-Clapeyron Equation Example

Enthalpy using Integrated Form of Clausius-Clapeyron Equation Solution

Enthalpy using Integrated Form of Clausius-Clapeyron Equation Formula Elements

Other formulas in Clausius Clapeyron Equation category

How to Evaluate Enthalpy using Integrated Form of Clausius-Clapeyron Equation?

FAQs on Enthalpy using Integrated Form of Clausius-Clapeyron Equation

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