How can i rearrange this Clausius-Clapeyron equation so that 1/T2 is on one side?
In P2/P1=delta H vaporization/R (1/T2- 1/T1)
I am trying to find at At what temperature does acetone have Pvap= 111 mm Hg.
Acetone, a common laboratory solvent, has Delta H vap= 29.1kJ/mol and a normal boiling point of 56.2 C.
I could easily do this problem if I was able to correctly rearrange the equation.
Can someone explain to me how to rearrange this?
Thank you very much!
To rearrange the Clausius-Clapeyron equation so that 1/T2 is on one side, you can follow the steps below:
1. Start with the original equation: P2/P1 = ΔH_vap/R (1/T2 - 1/T1).
2. Multiply both sides of the equation by (1/T2 - 1/T1) to isolate P2/P1: (P2/P1)(1/T2 - 1/T1) = ΔH_vap/R.
3. Distribute the multiplication on the left side: P2/T2 - P2/T1 = ΔH_vap/R.
4. Move P2/T1 to the other side of the equation by adding it to both sides: P2/T2 = ΔH_vap/R + P2/T1.
5. Subtract ΔH_vap/R from both sides to isolate P2/T2: P2/T2 - ΔH_vap/R = P2/T1.
6. Multiply both sides of the equation by T2 to isolate 1/T2: 1/T2 = P2/T1 - ΔH_vap/(R*T2).
By rearranging the Clausius-Clapeyron equation as shown above, you can solve for 1/T2 and calculate the temperature at which acetone has a vapor pressure (Pvap) of 111 mm Hg.
To rearrange the Clausius-Clapeyron equation so that 1/T2 is on one side, you need to isolate the term 1/T2.
Let's start with the original equation:
P2/P1 = ΔHvap/R * (1/T2 - 1/T1)
First, multiply both sides of the equation by (1/T2 - 1/T1):
P2/P1 * (1/T2 - 1/T1) = ΔHvap/R * (1/T2 - 1/T1) * (1/T2 - 1/T1)
Expand the right side:
P2/P1 * (1/T2 - 1/T1) = ΔHvap/R * (1/T2^2 - 2/T1T2 + 1/T1^2)
Then, distribute P2/P1 to both terms on the right side:
P2/P1 * (1/T2) - P2/P1 * (1/T1) = ΔHvap/R * (1/T2^2 - 2/T1T2 + 1/T1^2)
Next, rearrange the equation to isolate 1/T2:
P2/(P1*T2) - P2/(P1*T1) = ΔHvap/R * (1/T2^2 - 2/T1T2 + 1/T1^2)
Now, add P2/(P1*T1) to both sides:
P2/(P1*T2) = ΔHvap/R * (1/T2^2 - 2/T1T2 + 1/T1^2) + P2/(P1*T1)
Next, simplify the right side:
P2/(P1*T2) = ΔHvap/R * (1/T2^2 - 2/T1T2 + 1/T1^2) + P2/(P1*T1)
Finally, rearrange the equation by isolating 1/T2:
1/T2 = [ΔHvap/R * (1/T2^2 - 2/T1T2 + 1/T1^2) + P2/(P1*T1)] * (P1*T2/P2)
Simplify the right side:
1/T2 = ΔHvap/R * [1/(T2^2) - 2/(T1T2) + 1/(T1^2)] + T1/P2
Now, you have rearranged the Clausius-Clapeyron equation so that 1/T2 is on one side. You can substitute the given values (Pvap = 111 mm Hg, ΔHvap = 29.1 kJ/mol, T1 = boiling point) to solve for T2.