Question

Derive an expression for clapeyron latent heat equation

Answer 1

hii dear it is so long so i am giving link of this answer just touch it and get answer The Clausius-Clapeyron Equation: Its Derivation and Application
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Answer 2

Answer:

$\ln \frac{P_{2}}{P_{1}}=\frac{\Delta H_{v a p}}{R}\left(\frac{1}{T 1}-\frac{1}{T 2}\right)$ it is the Clausius clapeyron equation

Explanation:

• The Clausius Clapeyron equation predicts the rate at which vapor pressure increases per unit increase in temperature for a substance's vapor pressure (P) and temperature (T).

             $\frac{d l n p}{d T}=\frac{\Delta H_{v a p}}{R T_{2}}$

• The molar enthalpy of vaporization of the liquid, the ideal gas constant, and the temperature of the system determines the rate at which the natural logarithm of the vapor pressure of a liquid varies with temperature, according to this equation.
• If Hvar is assumed to be independent of the system's temperature, the Clausius Clapeyron equation can be written in the integrated form below, where C is a constant.
• Where Hvap is the liquid's enthalpy of evaporation, R is the gas constant, and A is a constant whose value is determined by the substance's chemical identity.
• In this equation, the temperature (T) must be in kelvin. Since the relationship between vapor pressure and temperature is not linear, the equation is often rewritten in the logarithmic form to provide the following linear equation:

              $\ln (P)=\frac{-\Delta H_{v a p}}{R T}+\ln A$

• If the enthalpy of evaporation and vapor pressure at a given temperature is defined for every liquid, the Clausius Clapeyron equation can be used to calculate the vapor pressure at a different temperature.
• The linear equation can be formulated in a two-point format to accomplish this. If the vapor pressure at temperature T1 is P1 and the vapor pressure at temperature T2
• is P2, the corresponding linear equations are:$\ln \left(P_{1}\right)=-\frac{\Delta H_{v a p}}{R T_{1}}+\ln A$ And

        $\ln \left(P_{2}\right)=-\frac{\Delta H_{v a p}}{R T_{2}}+\ln A$

• Since the constant, A, is the same in both equations, they can be rearranged to separate ln A and then made equal:
• $&\ln \left(P_{1}\right)=-\frac{\Delta H_{v a p}}{R T_{1}}+\ln A\\\ln \left(P_{2}\right)=-\frac{\Delta H_{v a p}}{R T_{2}}+\ln A \\&\ln \left(P_{1}\right)=-\frac{\Delta H_{v a p}}{R T_{1}}\\\ln \left(P_{2}\right)=-\frac{\Delta H_{v a p}}{R T_{2}}\end{aligned}$
• which can be combined into:
• $\ln \frac{P_{2}}{P_{1}}=\frac{\Delta H_{v a p}}{R}\left(\frac{1}{T 1}-\frac{1}{T 2}\right)$
• Hence it is the Clausius clapeyron equation.