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Worksheet 8 – Answers to Critical Thinking Questions The worksheets are available in the tutorials and form an integral part of the learning outcomes and experience for this unit. Model 1: The Gas Laws 1. Boyle’s Law: 2. Charles’ Law: 3. Avogadro’s Hypothesis: 4. (a) V = nRT ! ! k B = nRT (b) V = !" ! T k C = !" ! (c) V = !" ! n k A = !" ! 5. R = !" !" = !.!!! !"# ×(!!.!"! !) !.!!! !"# × (!"#.!" !) = 0.08206 atm L mol -1 K -1 The units arise directly from the equation: (atm × L) / (mol × K). 6. (a) 22.414 L corresponds to 0.022414 m 3 . (b) R = !" !" = !.!"#$% ×!" ! !" ×(!.!""#$# ! ! ) !.!!! !"# ×(!"#.!" !) = 8.314 Pa m 3 mol -1 K -1 The units arise directly from the equation: (Pa × m 3 ) / (mol × K). Model 2: Partial Pressures 7. (a) ! ! = 0.80 × 1.0000 atm = 0.80 atm } P total = (0.20 + 0.80) atm = 1.00 atm ! ! = 0.20 × 1.0000 atm = 0.20 atm (b) As density ρ = ! ! and m = nM, ρ = !" !" , the molar mass M = !" ! : P V T V n V

CHEM1901/3 Worksheet 8 – Answers to Critical Thinking Questions

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Worksheet 8 – Answers to Critical Thinking Questions The worksheets are available in the tutorials and form an integral part of the learning outcomes and experience for this unit. Model 1: The Gas Laws 1. Boyle’s Law: 2. Charles’ Law: 3. Avogadro’s Hypothesis:

4. (a) V = nRT !!

kB = nRT

(b) V = !"!

T kC = !"!

(c) V = !"!

n kA = !"!

5. R = !"!"= !.!!!  !"#  ×(!!.!"!  !)

!.!!!  !"#  ×  (!"#.!"  !) = 0.08206 atm L mol-1 K-1

The units arise directly from the equation: (atm × L) / (mol × K). 6. (a) 22.414 L corresponds to 0.022414 m3.

(b) R = !"!"= !.!"#$%  ×!"!  !"  ×(!.!""#$#  !!)

!.!!!  !"# ×(!"#.!"  !) = 8.314 Pa m3 mol-1 K-1

The units arise directly from the equation: (Pa × m3) / (mol × K).

Model 2: Partial Pressures 7. (a) 𝑃!!= 0.80 × 1.0000 atm = 0.80 atm

} Ptotal = (0.20 + 0.80) atm = 1.00 atm 𝑃!! = 0.20 × 1.0000 atm = 0.20 atm

(b) As density ρ = !!

and m = nM, ρ = !"!"

, the molar mass M = !"�!

:

P

V

T

V

n

V

M = !.!"#!$  !"#  !  !"#!!  !!! × !"#  ! ×(!.!"#  !  !!!)

(!.!!!  !"#) = 29.00 g mol-1

(c) 𝑀!! = 28.02 g mol-1 and 𝑀!!= 32.00 g mol-1. If x is the percentage by mass of O2 then: 29.00 = 32.00x + 28.02(1 - x) Solving this gives x = 0.25 and so air is 25% O2 and 75% N2 by mass.

Exercises 1. At 15.0 m, P = 2.50 atm. V15.0 m = 2.40 L. 2. At 30.0 m, P = 4.00 atm. Vsurface = 20. L. It will burst. 3. Air caught in a cavity will try to expand as the pressure is reduced during ascent. If trapped, it may cause

severe pain or a perforated eardrum in the ear or very severe toothache in a tooth. 4. P35 °C = 209 atm.

5. The increasing pressure leads to an increase in the density, ρ = !"!"

. More air is held in the same volume so the density increases.

6. From question 7 in Model 2, 𝑃!! = 0.20 atm at the surface. At a depth of 10.0 m, Ptotal = 2.0 atm and so

𝑃!! = 0.40 atm. The increase in total pressure does not affect the percentage composition of the air. 7. If 𝑃!! = 1.6 atm then Ptotal = 8.0 atm. This corresponds to a depth of 70.0 m.

Model 3: Osmotic pressure

1. See below.

2. See above. 3. The pressure is higher on the side with the higher water level.

4. (i) 0.35 M (ii) 0.15 M (iii) 0.50 M 5. 1.00 M = 1.00 mol L-1 = 1.0 × 103 mol m-3

6. π = (1.0 × 103 mol m-3) × (8.314 J K-1 mol-1) × (298 K) = 2.5 × 106 Pa = 24 atm. Pretty big! Model 4: Osmotic pressure in biology 1. See below

The Ideal Gas Law

(i) Hypotonic medium (iii) Hypertonic medium

2. (i) Increase. (ii) Stay the same. (iii) Decrease. 3. The cells in the blood will swell and possibly burst. This can be fatal.