The vapor pressure of a liquid doubles when the temperature is raised from 76° C to 84° C.?
Full Question:
The vapor pressure of a liquid doubles when the temperature is raised from 76°
C to 84°
C. At what temperature will the vapor pressure be seven times the value at 76°
C?
Update:Full Question:
The vapor pressure of a liquid doubles when the temperature is raised from 76°
C to 84°
C. At what temperature will the vapor pressure be seven times the value at 76°
C?
I've tired doing the following but its not right:
84-76 = 8
8/2=4
4*7=28
74+28 = 104
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Maybe you could do it without using the clapeyron equation.
(ADDITION: I should have taken into account that the number of moles changes. That makes the method I use wrong. Although it’s close to that found with the clapeyron equation. None the less the method is still wrong and I’d go with the clapeyron. If you want to bore yourself with what I did please continue reading )
Assuming the system is non-adiabatic and vapor behaves as a ideal gas, use the ideal gas law:
P(old) * V = nRT(old) and P(new)*V = nRT(new)
P(new) = P(old)*2
Subtracting the two gives delta P
deltaP = Pnew - Pold
= Pold*2 - Pold
=Pold*(2-1) = Pold*2
deltaP = Pold*2 = (nR/V)*deltaT
which makes you able to solve for Pold
Pold = (nR/V)*8/2
Do the same as above but for P(old)*7
P(old)*7 - P(old) = Pold*6 = nR/V (newDeltaT)
Plugging in for Pold and canceling nR/V gives
4*6 = newDetlaT = Tnew - Told = Tnew - 349.15 K
Solving for Tnew gives
Tnew = 373.15 K which is 100 C
did you try using the clausius clapeyron equation?
.. ln(P1/P2) = (-dHvap/R) * (1/T1 - 1/T2)
???
********
you have 3 data points
.. (T1, P1) = (349.15K, P1)
.. (T2. P2) = (357.15K, 2*P1)
.. (T3, P3) = (T3, 7*P1)
... .... .... .... ...↑
... ... ...... you're asked to find T3
********
hints
.. (1) write out 2 equations
.. .. ... equation #1, use the 2 data points (T1,P1) and (T2,P2)
... ... ..equation #2, use the 2 data points (T1,P1) and (T3,P3)
.. (2) divide to cancel out (-dHvap/R)
.. (3) plug and chug
********
********
********
solution
from the c.c. equation
.. ln(P1/P2) = (-dHvap/R) * (1/T1 - 1/T2)
.. ln(P1/P3) = (-dHvap/R) * (1/T1 - 1/T3)
dividing
.. ln(P1/P2) / ln(P1/P3) = (-dHvap/R) / (-dHvap/R) * (1/T1 - 1/T2) / (1/T1 - 1/T3)
canceling
.. ln(P1/P2) / ln(P1/P3) = (1/T1 - 1/T2) / (1/T1 - 1/T3)
plugging in just the P values for now
.. ln(P1 / 2*P1) / ln(P1 / 7*P1) = (1/T1 - 1/T2) / (1/T1 - 1/T3)
canceling
.. ln(1/2) / ln(1/7) = (1/T1 - 1/T2) / (1/T1 - 1/T3)
using the relationships ln(a/b) = ln(a) - ln(b) and ln(1) = 0
.. ln(2) / ln(7) = (1/T1 - 1/T2) / (1/T1 - 1/T3)
the rest is trivial
.. (1/T1 - 1/T3) = (1/T1 - 1/T2) * ln(7) / ln(2)
.. 1/T3 = (1/T1) - [(1/T1 - 1/T2) * ln(7) / ln(2)]
plugging in T data
.. 1/T3 = (1/349.15K) - [((1/349.15K) - (1/357.15K)) * ln(7)/ln(2)] = 0.002684/K
.. T3 = 1/0.002684/K = 372.58K = 99°C
Hmm it's normal