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dana.e
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Homework Statement
During a lab experiment my partner and I measured the average (taken from 6 measurements) time it took for 1 litter of liquid hellium to evaporate with and without adding heat to the system. We find out that without heating it took 54.16 seconds for 1 litter of hellium to evaporate, and with adding 100mW it took 40.16 seconds.
Our purpose is to find the lattent heat of hellium using the Clausius–Clapeyron relation.
We found that volume of 1.84*10^-5 m^3 evaporated in 1 second.
The temperate of the system was 3.71K and the pressure was 454 torr (= 60528.19 pascal). for dP/dT in the equation we took the temperature and pressure in that area- T2=3.7K, P2=450 torr (=59994.9 pascal) giving gradient of dP/dT= 53328.8 pascal/K.
Homework Equations
The relation of Clausius–Clapeyron is: dP/dT=L/(T*Vg) where dP/dT is the gradient calculated above, L is the lattent heat we want to find, T is the the system's temperature, and Vg is the volume of the Hellium gas which was evaporated per mass.
The lattent heat is also given by L=Q/m, where Q is the amount of heat that is used to change the state of the hellium from liquid to gas, and m is the mass.
The Attempt at a Solution
We calculated the number of moles in hellium gas using the connection PV=n*Kb*T, avogadro number (N0= 6.022*10^23) and the molar mass of hellium (M= 4.002602*10^-3 Kg) where P is the pressure, V is the volume, n is the number of moles, Kb is the constant Kb= 1.38*10^-23, finding that n=0.036, and the mass is 0.000144Kg.
By putting the measured values in the Clausius–Clapeyron relation with no heat added, we got an approximated value of L=25333J/Kg, where the theoretical value is 21000J/kg.
The major problem arised when we calculated the lattent heat with the addition of Q=100mW. From L=Q/m the power we added to the system is P=693.5W.
The value of the total lattent heat which was measure is 34163.85J/Kg, which is not correlating to the lattent heat we calculated when we didn't add heat to the system + the heat we added.
We wanted to know if our approach was right, what could have cause such big differences of the values we calculated, and if there is any other way to calcue the lattent heat using the Clausius–Clapeyron relation.
Thanks,
Dana.