we do not change the peessure or the temperature (see Figure 8.7). A and senpera

Question

we do not change the peessure or the temperature (see Figure 8.7). A and senperature, we san wite Avogadro's law s Law eROBLEM 8.7 Calculating Volume for a Change in assignmentProblemID-86356929 int Resources « previous | 4 of 7 next Nothing will happen O The water will catch on fire O The water will disappear instantly Submit My Answers Give Up Part C A sealed glass jar has a small amount of water ón the bottom. The rest of the volume in the jar is taken up by helium gas at a partial pressure of 526 mmlg. and water vapor, If the total pressure inside the jar is 760, mmHg, what is the temperature of the water? Express your answer to three significant figures and include the appropriate units. Hints rValveUnits Submit My Answers Give U Continue

Explanation / Answer

We know that;

Ptotal = PHe + PH2O

given that ;

Ptotal = total pressure ;760 mmHg
PHe = partial pressure helium; 526 mmHg

Then

PH2O = 760 – 526 mmHg

= 234 mmHg

now we take pressure of water vapor at its boiling point, 760 mmHg at 100 C or 373.165 K

And use clausius clapeyron equation.

ln(P1/P2) = (dHvap / R) x (1/T2 - 1/T1)

where.
P1 = 234mmHg
P2 = 760mmHg
dHvap = 40.68 kJ/mole = 40680 J/mole

R = 8.314 J/moleK
T2 = the corresponding Temperature of P2 = 100°C = 373.15K
T1 = ?

therefore;

ln(234/760) = ((40680 J/mole) / (8.314 J/moleK)) x (1/ 373.15 - 1/T1)

ln 0.31 = 4892.95 (1/ 373.15 - 1/T1)

-1.17 /4892.95 = 2.68*10^-3 – 1/T1

- 2.39*10^-4- 2.68*10^-3 = – 1/T1

1/T1 = 0.00292
T1 = 342.46 K

= 69.3°C

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