Using the Clausius Clapeyron equation determine the vapor pressure of water at 50.0 degrees C. The molar heat of vaporization of water is 40.7 kJ/mol. How do you solve this?
To solve this problem, we can use the Clausius-Clapeyron equation, which relates the vapor pressure of a substance to its temperature and molar heat of vaporization. The Clausius-Clapeyron equation is given as:
ln(P2/P1) = (-ΔHvap/R) * (1/T2 - 1/T1)
where P1 and P2 are the vapor pressures at temperatures T1 and T2 respectively, ΔHvap is the molar heat of vaporization, R is the ideal gas constant, and ln represents the natural logarithm.
In this case, the temperature T1 is the normal boiling point of water, which is 100.0 degrees Celsius (373.15 K). The temperature T2 is 50.0 degrees Celsius (323.15 K). The molar heat of vaporization of water, ΔHvap, is provided as 40.7 kJ/mol.
Now, let's go ahead and solve the problem step by step:
1. Convert the molar heat of vaporization, ΔHvap, from kilojoules to joules:
ΔHvap = 40.7 kJ/mol * 1000 J/1 kJ = 40,700 J/mol
2. Plug in the values into the Clausius-Clapeyron equation:
ln(P2/P1) = (-40,700 J/mol / R) * (1/323.15 K - 1/373.15 K)
3. Determine the value of the ideal gas constant, R. The value of R depends on the units used for pressure and volume. In this case, we are using the molar heat of vaporization in joules, so we'll use the value for R in J/(mol K), which is approximately 8.314 J/(mol K).
4. Substitute the value of R into the equation:
ln(P2/P1) = (-40,700 J/mol / 8.314 J/(mol K)) * (1/323.15 K - 1/373.15 K)
5. Simplify the equation and solve for ln(P2/P1):
ln(P2/P1) = -4.9149 * (0.003093 - 0.002678)
6. Calculate the difference in the logarithms and solve for P2/P1:
ln(P2/P1) = -4.9149 * (0.000415)
7. Calculate the value of P2/P1 using the natural logarithm, e:
P2/P1 = e^(-4.9149 * 0.000415)
8. Finally, solve for P2 by multiplying both sides of the equation by P1:
P2 = P1 * e^(-4.9149 * 0.000415)
Using this process, you can calculate the vapor pressure of water at 50.0 degrees Celsius.
Mg and O lab
how will the relative number of moles of mg and o be affected if the following procedural errors are made?
a)crucible is not brought to constant mass before the Mg is added.
b)water is not added after the first heating of Mg, and the entire solid in the crucible is considered to be MgO
c)some of the fluffy white ppt is lost before the crucible is weighed for the last time.
d)in an effort to begin a second sample while the first is cooling, the hot crucible containing the solid from determination 1 is accidently dropped, and a small chip is lost from the crucible.
i said b was a higher value for O because the mass of our product would be higher and we determined the mass of O by subtracting our mass of Mg used, in a previous calculation
i said the decrease in the amount of O for part c because of basically the same reason; we have a smaller product so we would calculate a smaller amount of O
the other two i'm not sure of and am kind of confused.
if you could check mine and help me with the others?
please be specific!thanks!