Question

Barometric Pressure (mmHg) Data Table (1 pt. each blank) 767 mmHg (1 pt.) Temperature of Water (°C) Volume of Air Bubble (mL) 2.4°C 3.00mh 50.3°C 3.80ml 55.0°C 4.00ml 60.0°C 4.30 mr 65.0°C 4.80mL 70.0°C 5.10mL 75.0°C 5.80mL show you mo42 80.5°C 6.90 mL Calculations: (0.5 pt for each block) (x) (y) Twater (K) 1/Twater (K?¹) Pair (mm Hg) Pwater (mm Hg) Pwater (atm) In Pwater 323.65 K 0.003 K?¹ 328-15 K 0.0034?¹ 333.15 K 0.003K 338.15K 0.003k 343.15K 0.003K?¹ 348.15K 0.003?¹ 353.65 K 0.003 ? P = Pair (2.4?) = Pbar (767mmHg) Vi = 3.00mL Ti = 275.55K

          Barometric Pressure (mmHg)
Data Table (1 pt. each blank)
767 mmHg (1 pt.)
Temperature of Water (°C) Volume of Air Bubble (mL)
2.4°C
3.00mh
50.3°C
3.80ml
55.0°C
4.00ml
60.0°C
4.30 mr
65.0°C
4.80mL
70.0°C
5.10mL
75.0°C
5.80mL
show you mo42
80.5°C
6.90 mL
Calculations: (0.5 pt for each block)
(x)
(y)
Twater (K)
1/Twater (K?¹) Pair (mm Hg) Pwater (mm Hg)
Pwater (atm)
In Pwater
323.65 K
0.003 K?¹
328-15 K
0.0034?¹
333.15 K
0.003K
338.15K 0.003k
343.15K 0.003K?¹
348.15K 0.003?¹
353.65 K 0.003 ?
P = Pair (2.4?) = Pbar (767mmHg)
Vi = 3.00mL
Ti = 275.55K

Show more…

Barometric Pressure (mmHg)
Data Table (1 pt. each blank)
767 mmHg (1 pt.)
Temperature of Water (°C) Volume of Air Bubble (mL)
2.4°C
3.00mh
50.3°C
3.80ml
55.0°C
4.00ml
60.0°C
4.30 mr
65.0°C
4.80mL
70.0°C
5.10mL
75.0°C
5.80mL
show you mo42
80.5°C
6.90 mL
Calculations: (0.5 pt for each block)
(x)
(y)
Twater (K)
1/Twater (K?¹) Pair (mm Hg) Pwater (mm Hg)
Pwater (atm)
In Pwater
323.65 K
0.003 K?¹
328-15 K
0.0034?¹
333.15 K
0.003K
338.15K 0.003k
343.15K 0.003K?¹
348.15K 0.003?¹
353.65 K 0.003 ?
P = Pair (2.4?) = Pbar (767mmHg)
Vi = 3.00mL
Ti = 275.55K

Added by Sheryl -.

Chemistry: Structure and Properties

Nivaldo Tro 2nd Edition

Instant Answer

Solved by Expert Kartik Indoliya

12/04/2021

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Data Table (1 pt. each blank) Calculations: for each block Barometric Pressure (mmHg) Data Table (1 pt. each blank) Hot Hg (1 pt.) Temperature of Water (°C) 2.4°C 50.5°C 7.0°C 56.0°C 65.0°C 0.00°C 2.0°C 80.5°C Volume of Air Bubble (mL) 3.00 mL 3.80 mL 4.00 mL 4.30 mL 103.0 mL 5.10 mL 5.80 mL 6.90 mL Calculations: (0.5 pt for each block) (x) Twater (K) 1/Twater (K^-1) Pair (mm Hg) (y) ln Pwater Pwater (mm Hg) Pwater (atm) 323.65 K 0.003 K^-1 328.15 K 0.003 K^-1 333.15 K 0.003 K^-1 338.15 K 0.003 K^-1 343.15 K 0.003 K^-1 348.15 K 0.003 K^-1 353.65 K 0.003 K^-1 P = Pwater = PBav = 767 mmHg V = 3.00 mL T = 275.55 K

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Transcript

00:01 First of all, let's have a look what all is given in the problem.

00:07 Given is moles of air molecule are given.

00:12 Moles of air molecules in the flask are given as 8 .908 into 10 kipar minus 4.

00:20 And a table is given which shows temperature and pressure reading.

00:25 And we need to complete the given table.

00:28 We need to complete the given table.

00:32 We need to find pressure from air molecules and vapor pressure from water.

00:37 We per pressure of water.

00:39 So we need to complete the given table.

00:44 We know that according to ideal gas equation, pvi equals to nirt.

00:51 So, p of air, p of air will be equal to n of air into rt divided by v.

01:01 Let's put the value.

01:02 Number of moles is given as 8 .90.

01:05 0 .08 into 10 kx minus 4 into gas constant is 8 .314 and the temperature is 321 .2 kelvin divided by volume is 1 litre.

01:17 So if we calculate this, it will come around 2 .379 kilo -pascal.

01:23 This is the pressure of air.

01:26 Now we know that p total will be equal to pressure of air plus pressure of vapor.

01:32 So from here we can calculate the pressure of vapor...

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