These are some lab questions from the following lab: Vapor Pressure and Enthalpy of Vaporization of Water
1. Why does vapor pressure change with temperature?
2. The assumption was made that the vapor pressure of water is negligible at a temperature close to zero. Find the actual vapor pressure of water at you low temperature and comment on the validity of the assumption.
(what equations/calculations do I use?)
3. The assumption was also made that the slight changes in "h", the depth under the surface of the water, will not significantly change the total pressure in the graduation cylinder. Comment on the validity of this assumption.
4. Write out the long "two-point" form of the Clausius-Clapeyron equation. Why does the graphical of analysis give a better value for the enthalpy of vaportization than does the form of the Clausius-Clapeyron equation using two temperature-vapor pressure values?
Please help! Thanks !!
1. Why does vapor pressure change with temperature?
Think kinetic molecular theory. Look up the KMT and explain from there.
2. The assumption was made that the vapor pressure of water is negligible at a temperature close to zero. Find the actual vapor pressure of water at you low temperature and comment on the validity of the assumption.
(what equations/calculations do I use?)
I'm not sure what you did; however, I think the question is asking you to measure the vapor pressure at the lowest temperature at which the experiment was conducted. That probably means you have the number somewhere in the data you collected. If that isn't correct, then I would interpret it to mean that you look up the value of the vapor pressure in a set of tables.
3. The assumption was also made that the slight changes in "h", the depth under the surface of the water, will not significantly change the total pressure in the graduation cylinder. Comment on the validity of this assumption. You can calculate the pressure of a column of water h cm high and that will allow you to compare with the pressures you used.
4. Write out the long "two-point" form of the Clausius-Clapeyron equation. Why does the graphical of analysis give a better value for the enthalpy of vaportization than does the form of the Clausius-Clapeyron equation using two temperature-vapor pressure values?
My take on this is that the graphical solution provides a smoothing out effect over many points. With a two-point solution, you have a straight line, period.
thankyou !!
hey--we just did this lab! and ahhh! i'm still clueless about the two point clausius-clapeyron eq. anyone have any more help?
We just did this same lab also and it was so confusing, but you helped just a little bit hee hee internet is great!!! :)
can you answer this??
what are the 2 assumptions of KMT that maybe considered vaque
Two- point equation:
ln (P1/P-2) = - (ÄHvap/R)(1/T2-1/T1)
The graphical analysis gives a more exact value because it takes into account possibly erroneous data points, making an “average”, “trend” or “best fit” line that is as close as possible to all data points. The equation method only utilizes two of the data points, which may or may not be representative of the rest of the data, and which may or may not be skewed.
thank you!! i was stuck on the last question but this helped :))
I'm glad I could provide some assistance! Let's dive a bit deeper into the Clausius-Clapeyron equation and explain the two-point form.
The Clausius-Clapeyron equation relates the vapor pressure of a substance to its temperature and heat of vaporization. The general form of the equation is:
ln(P2/P1) = -(ΔH_vap/R)(1/T2 - 1/T1)
where P1 and T1 are the initial pressure and temperature, P2 and T2 are the final pressure and temperature, ΔH_vap is the enthalpy of vaporization, and R is the gas constant.
Now, the two-point form of the equation is obtained by arranging the equation and solving for ΔH_vap. It takes the following form:
ΔH_vap = -((R*T1*T2)/(T2 - T1))*ln(P2/P1)
To calculate the enthalpy of vaporization using this equation, you need to have two sets of temperature and vapor pressure values. Then you can plug those values into the equation to obtain the enthalpy of vaporization.
Now, the question is asking why the graphical analysis gives a better value for the enthalpy of vaporization compared to the two-point form of the equation.
When you create a graph of ln(P) vs. 1/T using multiple data points, you can determine the slope of the line and calculate the enthalpy of vaporization using the equation:
ΔH_vap = -slope*R
The advantage of the graphical analysis is that it utilizes multiple data points, resulting in a more accurate representation of the relationship between vapor pressure and temperature. It considers the curvature and variations in the data, which can better account for any experimental errors or deviations from the ideal gas behavior.
On the other hand, the two-point equation only considers data at two specific temperature and pressure points, leading to a simplified straight-line relationship. This simplification may not accurately capture the true behavior of the substance and can be more influenced by errors or uncertainties in the experimental data.
That's why the graphical analysis generally provides a better value for the enthalpy of vaporization compared to the two-point form of the Clausius-Clapeyron equation.