Grading Rubric: Charles Law Lab
Only the questions listed were graded. Please check your answers with these and let me know of any corrections as soon as possible.
#5. 2pts. Velocity is high, but mass is low, so KE is low.
#8. 4pts. a. 760mm – 23.7mm = 736.24mmHg is gas pressure alone 98.16 Kpa.
V2 = 48.43mL
#10. 2pts. Part I: C, because no air escaped.
Part II: A, because molecules will speed up and increase KE and pressure.
#11. 2pts. Part I: A, because air escaped.
Part II: C, because it is open to the air and it goes into equilibrium with the air.
* There should be 3 data tables.
Solve for This
#1. 5pts. (V2 uncorrected)(P room – PH2O) = (P room)(V2corrected) ¿
This should be done for all 3 data tables. V2corrected should be less that V2uncorrected.
Solve for This
#2. 5pts. (V2corrected)(T1) = (T2)(V1 experiment) ¿
This should be done for all 3 data tables. And V1experiment should be very close (but not exactly) equal to V1.
Post Lab Questions:
#2. 6pts. A. PV1experiment = nRT2 ¿ (Should be more moles than 2b and C)
B. PV1experiment = nRT1¿
Room These should be similar in Value for “n”
C. PV 2 corrected = nRT2¿
#4. 2pts.That ensures the Pressure in the flask will be equal to the room pressure. If the water in the flask were higher, the P would be lower. If the water in the flask was lower, the P would be greater.
#5. 1pt. Low Pressure flask is replaced by water forced in by P room.
#6. 1pt. Should be near –273 oC.
10 points for the graph
10 points for neatness