5 If water boils at 273 K at sea level (pressure 1 atm), what will be the temper

Question

5 If water boils at 273 K at sea level (pressure 1 atm), what will be the temperature at which the water boils if the pressure is reduced to 0.93 atm? A. 314 K B. 254 K C. 238 K D. 273 K 5 If water boils at 273 K at sea level (pressure 1 atm), what will be the temperature at which the water boils if the pressure is reduced to 0.93 atm? A. 314 K B. 254 K C. 238 K D. 273 K 5 If water boils at 273 K at sea level (pressure 1 atm), what will be the temperature at which the water boils if the pressure is reduced to 0.93 atm? A. 314 K B. 254 K C. 238 K D. 273 K

Explanation / Answer

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)

P1 = 1 atm, T = 373 K CORRECTION, WATER DOES NOT BOILS AT 273 K AT P = 1 ATM, IT BOILS AT 100°c OR 373 K

dHvap = 40.65 = 40650 J/mol

ln(0.93/1) = 40650 /8.314*(1/(373) - 1/(T2))

ln(0.93/1) /40650 * 8.314 - 1/(373) = -1/T2

T2= -(-0.0026958)^-1

T2 = 370K

CLEARLY, the error is included in the answer, so assume T = 273 K

ln(0.93/1) /40650 * 8.314 - 1/(273) = -1/T2

T2= -(-0.003677)^-1

T2 = 272 K, nearest answer is D

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