Question

Please help with questions 1-7!!!!

8.79 for larm 29. GainHg : 760 torr latm IV. DATA Name Enima taub Trial 2 Trial 1 Volume of flask with stopper 1353. ImL 329 UL Volume of water in flask 70.0 ml Volume of the cool gas 283. I ml 260ML Temperature of boiling water 100.0'C→ 373.6K 49.66 | 372.6 Temperature of the cool water 19.3°C 282.9k 18.5" 281.54 Barometric pressure 29.01 in Hg 29.06InHg Vapor pressure of water 8.79 Torr or 0.34 binty 8.32 Torr Partial pressure of cool gas 28.7in Hs 28.732a Hy Lab Partner Aya Soblani Data Source for Trial 2 Aishe Mubarak NOA Notes: or the general sia include Celsius mus. For exampl 1. The data table should contain data in the units actually measured. Although you will need Kelvin temperatures for the general gas law, the measurements are made with a Celsius thermometer. So the data table should include Celsius temperatures. It would acceptable to include the Kelvin temperature in the table with the Celsius. For example, the table might read: 18.0 °C -291.2K 2. Values that are lined up in the table can be subtracted directly as data. For example, the vapor pressure can be subtracted from the barometric above it to get the pressure of the cool gas, of course the units in all three boxes must be the same. One cannot subtract tort from em Hgor inches Hg without first converting to similar units. 3. Fach experimenter is responsible to do the subtractions correctly even when data is shared.

Experiment 12 The General Gas Law

Answer 1

There are many subparts in this question. I am answering part 1-4 (Chegg Policy: 1 questions per post or 4 sub parts in 1 question)

ANSWER:

Question 1: V2 for trial 1

Data:

• V1 = volume of flask = 353.1 mL
• T1 = temp of boiling water = 100 ºC = 373.2 K
• P1 = barometric pressure = 29.06 inHg
• T2 = temp of cool water = 9.3 ºC = 282.5 K
• vapor pressure of water at 9.3 ºC = 8.79 torr = 0.346 inHg

linHg 8.79 torr = 8.79 mm Hg = 8.79 mm Hg x= 925.4 mm Hg = 0.346 in Hg

• P2 = partial pressure of cool gas = barometric pressure - vapor pressure of water (at T2) = 29.06 inHg - 0.346 inHg = 28.714 inHg

Calculation:

To calulate V2, we use the combined gas law:

” v ()(会)

V2 = 29.06 in Hg x 353.1 mL 373.2 K 282.5 K 28.714 in Hg

V2= 270.51 mL

------------------------------------------------------------------------

Question 2: percent difference between V2 calculated and V2 experimental for trial 1

The percent difference is calculated as

% difference = V2 erp – V2 cale wel x 100% Vacal

% difference = 283.1 mL - 270.51 mL. 1x 100% = 4.65% 270.54 mL

------------------------------------------------------------------------------

Question 3: V2 for trial 2

Data:

• V1 = volume of flask = 329 mL
• T1 = temp of boiling water = 99.6 ºC = 372.8 K
• P1 = barometric pressure = 29.06 inHg
• T2 = temp of cool water = 8.5 ºC = 281.7 K
• vapor pressure of water at 8.5 ºC = 8.32 torr = 0.328 inHg

linHg 8.32 torr = 8.32 mm Hg = 8.32 mmHg x 925.4 mm Hg = 0.328 in Hg

• P2 = partial pressure of cool gas = barometric pressure - vapor pressure of water (at T2) = 29.06 inHg - 0.328 inHg = 28.732 inHg

Calculation:

To calulate V2, we use the combined gas law:

” v ()(会)

V2 = 29.06 in Hg x 329 mL 372.8 K | 281.7 K 28.732 in Hg

V2= 251.44 mL

------------------------------------------------------------------------

Question 4: percent difference between V2 calculated and V2 experimental for trial 2

The percent difference is calculated as

% difference = V2 erp – V2 cale wel x 100% Vacal

% difference = 260 mL - 251.44 mL... x 100% = 3.40% 251.44 mL