The instruction booklet for your pressure cooker indicates that its highest sett
ID: 633629 • Letter: T
Question
The instruction booklet for your pressure cooker indicates that its highest setting is 13.1psi . You know that standard atmospheric pressure is 14.7 psi, so the booklet must mean 13.1 psi above atmospheric pressure. At what temperature in degrees Celsius will your food cook in this pressure cooker set on "high"? Express your answer numerically in degrees Celsius. Which of the following factors affect the vapor pressure of a liquid? Check all that apply. addition of solute type of liquid surface area of the liquid amount of liquid temperature atmospheric pressureExplanation / Answer
The instruction booklet for your pressure cooker indicates that its highest setting is 11.2psi .
You know that standard atmospheric pressure is 14.7 psi, so the booklet must mean 11.2psi above atmospheric pressure.
At what temperature in degrees Celsius will your food cook in this pressure cooker set on "high"?
Recall that in equilibrium; especially in vapor-liquid equilibriums, we can use Clasius Clapyeron combination equation in order to relate two points in the same equilibrium line.
The equation is given as:
ln(P2/P1) = -dHvap/R*(1/T2-1/T1)
Where
P2,P1 = vapor pressure at point 1 and 2
dH = Enthalpy of vaporization, typically reported in kJ/mol, but we need to use J/mol
R = 8.314 J/mol K
T1,T2 = Saturation temperature at point 1 and 2
Therefore, we need at least 4 variables in order to solve this.
Substitute all known data:
ln(P2/P1) = -dHvap/R*(1/T2-1/T1)
Change negative signs
ln(P2/P1) = dHvap/R*(1/T1-1/T2)
we will ned
dHvap for water = 40.67 kJ/mol
substitute data:
T1 = 373 K, T2 = x
P1 = 14.7 psi; P2 = 11.2+14.7 = 25.9 psi
substitute
ln(P2/P1) = dHvap/R*(1/T1-1/T2)
ln(25.9 /14.7 ) = 40670/8.314*(1/373-1/T2)
T2 = -( ln(25.9 /14.7 )*8.314/40670 - 1/373)^-1
T2 = 389.836K
T2 = 389.836-273
T2 = 116.836 °C
16.83° higher!
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