| DATA / RESULTS | ||||
|---|---|---|---|---|
| TRIAL ONE | TRIAL TWO | |||
| Item | Height of Column | Temperature °C | Height of Column | Temperature °C |
| Hot water, cm | ||||
| 60° water, cm | ||||
| 30° water, cm | ||||
| Ice-water temperature, cm | ||||
| Length of mercury plug in mm at room temperature | ||||
| Atmospheric pressure, mm Hg; (See Barometer in Balance Room) | ||||
| Inside diameter of tube with plug of mercury, cm | ||||
| Experimental value of absolute zero | ||||
| Slope of the line (include units) | ||||
PV = nRT and V = (nR/P) T Where V= ha for a cylinder (h= height in cm and a = cross sectional area in cm). V is in cm3. Then ha = (nR/P) T or h = (nR/Pa) T
The slope of the line (when h is plotted vs T from above equation), multiplied by the cross-sectional area of the air column in dm2 (One liter is equal to one dm3) is equal to nR/P where R, is 0.08206 L-atm/K-mol. Note the pressure must be in units of atm.