Finding the Lattent heat of Hellium - a lab experiment.

[dana.e]

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.

 

[mark726]

Dear Dana,

Thank you for sharing your experiment and calculations with us. It seems like you have taken a good approach to finding the lattent heat of helium using the Clausius–Clapeyron relation. However, there are a few things that could be causing the discrepancies in your results.

Firstly, it is important to make sure that all your measurements are accurate. Small errors in measurement can lead to significant differences in the final results. Double-check your measurements and calculations to ensure they are correct.

Another factor to consider is the experimental setup. Was the system completely closed and insulated from outside influences? Any external heat or pressure changes could affect the results. It would also be helpful to know the specific method you used to add heat to the system. Was it a steady and controlled addition of heat or was it applied all at once? This could also impact the results.

Additionally, it is important to note that the Clausius–Clapeyron relation is an approximation and may not give exact results. There could be other factors at play that are not accounted for in the equation.

One way to validate your results could be to compare them with other sources or previous experiments on the lattent heat of helium. This can help identify any potential errors or discrepancies in your experiment.

Overall, your approach seems reasonable and it is possible that the discrepancies in your results could be due to experimental errors. I would recommend reviewing your measurements and experimental setup to ensure accuracy, and also comparing your results with other sources. I hope this helps and good luck with your experiment!