Clausius Clapeyron equation

In summary, the problem involves finding the heat of vaporization (Hvap) using the Clausius Clapeyron equation, which is ln P= (-delta Hvap/R)*(1/T) + C. The given slope of the graph is -8000 K, which corresponds to the slope of the equation y=mx + b. By rearranging the equation, it is possible to solve for -deltaHvap and find the value for Hvap.
  • #1
mygymnst
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Homework Statement



slope: -8.000 x 10^3 K
The graph is of ln vapor pressure of a gas vs. invert kelvin temperature
Using the Clausius Clapeyron equation, find Hvap (heat of vaporization)

Homework Equations



Clausius Clapeyron equation: ln P= (-delta Hvap/R)*(1/T) + C

The Attempt at a Solution



I tried using the given slope in the equation as T. Then I tried using y=mb. I multiplied the slope by 100K thinking I could find lnP.
 
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  • #2
In this example you have the equation in the form of a line already.

y=mx + b

where y is lnP, m is -deltaHvap/R, x is 1/T and b is C. You are asked to determine the value of -deltaHvap given that the slope (m) is -8000 K.

Can you find a way to tease out the -deltaHvap from the expression for the slope (-8000 K = -deltaHvap/R)?
 
  • #3
I also tried converting the slope to lnP by taking the natural log of the slope. However, I am not sure how to use the given information to find Hvap.

The Clausius Clapeyron equation is a fundamental relationship in thermodynamics that relates the vapor pressure of a substance to its temperature and heat of vaporization. In this case, the given slope of -8.000 x 10^3 K represents the change in temperature (T) over the change in ln vapor pressure (ln P). To solve for Hvap, we can rearrange the equation to isolate for Hvap:

Hvap = -R*(slope)*(1/T) + C

Where R is the gas constant and C is the y-intercept of the graph.

To find the heat of vaporization, we need to plug in the values for slope, temperature, and gas constant into the equation. The given temperature is in Kelvin, so we do not need to convert it. However, we need to convert the slope from K to Kelvin^-1 by dividing it by 1000, since the slope represents the change in temperature over 1000 units of ln vapor pressure.

Hvap = -(8.000 x 10^3 K)/1000 * (1/T) + C

Next, we need to determine the value of ln P at a specific temperature. We can do this by using the given information of the graph, which shows ln P on the y-axis and 1/T on the x-axis. We can choose a specific point on the graph, such as (1/T = 0.001, ln P = -5), and plug these values into the equation to solve for C.

C = ln P + (delta Hvap/R)*(1/T)
C = -5 + (8.000 x 10^3 K)/1000 * (0.001)
C = -5 + 8 = 3

Now, we can plug in the values for slope, temperature, gas constant, and C into the equation to solve for Hvap.

Hvap = -8.000 x 10^3 K/Kelvin * (1/0.001 Kelvin^-1) + 3

Hvap = -8.000 x 10^3 J/mol + 3

Hvap = -7.997 x 10^3 J/mol

Therefore, the heat of vaporization for
 

1. What is the Clausius Clapeyron equation?

The Clausius Clapeyron equation is a thermodynamic equation that relates the change in vapor pressure of a substance to its temperature. It is often used to calculate the vapor pressure of a substance at different temperatures, and can also be used to estimate the boiling point of a substance at a given pressure.

2. Who developed the Clausius Clapeyron equation?

The Clausius Clapeyron equation was developed independently by two scientists, Rudolf Clausius and Benoit Paul Émile Clapeyron, in the mid-19th century. Clausius published his version in 1850, while Clapeyron published his in 1834.

3. What are the assumptions made in the Clausius Clapeyron equation?

The Clausius Clapeyron equation assumes that the substance being studied is in equilibrium between its liquid and vapor phases, and that the vapor behaves as an ideal gas. It also assumes that the change in enthalpy and entropy between the liquid and vapor phases is constant over the temperature range being studied.

4. How is the Clausius Clapeyron equation used in practical applications?

The Clausius Clapeyron equation is used in various practical applications, such as weather forecasting, predicting the behavior of gases, and understanding phase transitions in materials. It is also used in the design and operation of refrigeration and air conditioning systems.

5. What are some limitations of the Clausius Clapeyron equation?

The Clausius Clapeyron equation is based on certain assumptions and is only accurate for a limited range of temperatures and pressures. It also does not take into account factors such as non-ideal behavior of gases and the presence of other substances in the system. Additionally, it may not accurately predict the behavior of substances with complex molecular structures or those undergoing phase transitions at very low temperatures.

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