Chapter 05: Gases

Question 1

Pressure

Answer 1

The force exerted per unit area by gas molecules as they strike the surfaces around them

Pressure = force/area = F/A

Question 2

Pressure units & conversion factors

Answer 2

760 mmHg = 760 torr = 1 atm = 101,325 Pa

Question 3

Elements that exist as gases at 25°C and 1 atm

Answer 3

1A: H

5A: N

6A: O

7A: F, Cl

8A: (all)

Question 4

Force (formula)

Answer 4

mass × acceleration

Question 5

SI units of force and pressure

Answer 5

Force: 1 newton (N)

Pressure: 1 pascal (Pa)

Question 6

Newton

Answer 6

1 newton = 1 kg×m/s²

Question 7

Pascal

Answer 7

1 pascal = 1N/m² = 1 kg/(m×s²)

Question 8

Barometer

Answer 8

Atmospheric pressure measurment tool

Inverted tube of Hg over open dish of Hg;
height of Hg in mm is equal to atmospheric pressure

Question 9

Manometer

Answer 9

Instrument used to measure pressure of gas trapped in container

Gas pressure is determined by difference in liquid levels in U-shaped tube

(Pressure is high if it pushes down on liquid;
low if it cannot push down on liquid)

Question 10

Simple Gas Laws: Four* basic properties of a gas

Answer 10

P, pressure

V, volume

T, temperature (Kelvin)

t, temperature (°C)

n, amount in moles

Question 11

Boyle’s law

Answer 11

Inverse relationship between volume and pressure

*n is constant
*T is constant

P₁V₁ = P₂V₂

Question 12

Charles’ law

Answer 12

Direct relationship between volume and temperature (Kelvin).

*P is constant
*n is constant

V₁ = V₂
T₁ T₂

Question 13

Avogadro’s law

Answer 13

Direct relationship between volume and quantity (number of moles)

*P is constant
*T is constant

V₁ = V₂
n n

Question 14

Ideal Gas Law

Answer 14

How a hypothetical gas behaves

PV = nRT

Where R is the ideal gas constant

Question 15

R, ideal gas constant

Answer 15

R = 0.08206 L atm/mol K

Question 16

Molar volume

Answer 16

The volume occupied by one mole of a substance

Question 17

Ideal gas molar volume

Answer 17

22.414 L at STP of an ideal gas

Question 18

STP

Answer 18

standard temperature and pressure

pressure = 1 atm at STP

temperature = 0°C or 273.15 K at STP

Question 19

Density of a gas (formula)

Answer 19

d = PM
RT

Where d = density
P = pressure in atm
M = molar mass
R = gas constant
T = temperature in Kelvin

Question 20

Molar mass of a gas (formula)

Answer 20

M = dRT
P

Where M = molar mass
d = density
R = gas constant
T = temperature in Kelvin
P = pressure in atm

Question 21

Dalton’s law

Answer 21

Partial pressures, P_n

P_n = n_nRT
V

P_total = P_x + P_y + P_z

*volume and temperature are constant

Question 22

Mole fraction

Answer 22

X_n = n_n
n_total

Thus, with Dalton’s law:
P_n = X_nP_total

Question 23

Vapor pressure

Answer 23

The partial pressure of water vapor in a system

Question 24

Gas stoichiometry

Answer 24

General conceptual plan:

P, V, T / mass or volume of gas A –> moles of gas A –>
moles of gas B –> P, V, T / mass or volume of gas B

Question 25

Kinetic molecular theory

Answer 25

1. Particle **size** is **negligible**, even those they have mass 2. Constant motion, random directions, **no overall loss of energy, just a transfer of energy** (known as being **_elastic_**) 3. **Average** **kinetic** **energy** is **directly** proportional to **temperature** in **Kelvin** 4. Gas particles exert **neither** **attractive** nor **repulsive** forces

Question 26

Average kinetic energy

Answer 26

average kinetic energy = KE ( KE = 1/2×mv² ) At **same** **temperature**, gases have **same** **average** **kinetic** **energy**

Question 27

Boyle's law explained (KMT)

Answer 27

Decreasing volume forces gas particles into a smaller space, thus increasing collisions, and hence, pressure \*n and T are constant

Question 28

Charle's law explained (KMT)

Answer 28

Temperature increase increases average kinetic energy The higher the average kinetic energy, the greater the area particles will move around Thus, volume increases \*P & n constant

Question 29

Avogadro's law explained (KMT)

Answer 29

Increasing number of particles causes more collisions To keep pressure constant, volume must then increase \*P and T are constant

Question 30

Dalton's law explained (KMT)

Answer 30

Because average kinetic energy is the same (at same temperature), the total pressure of the collisions is the same

Question 31

Root mean square velocity

Answer 31

A kind of average velocity root mean square velocity = u_rms u_rms = (3RT/*M*)^1/2 Where *M* = molar mass in **kg/mol** R = gas constant in **J/mol K** T = temperature in Kelvin

Question 32

R, gas constant (with joules)

Answer 32

8.314 J/mol K \*used as R in root mean square velocity

Question 33

1 Joule = ? (kg, m, s)

Answer 33

1 J = 1 kg×m²/s²

Question 34

Mean free path

Answer 34

The average distance a gas molecule travels between collisions mean free path _v as pressure ^

Question 35

Diffusion

Answer 35

Process by which gas molecules spread out in response to concentration gradient Heavier molecules diffuse more slowly than lighter ones

Question 36

Effusion

Answer 36

The process by which a gas escapes from a container into a vacuum through a small hole Heavier molecules effuse more slowly than lighter ones

Question 37

Rate of effusion

Answer 37

Amount of gas that effuses in a given time Inversely proportional to square root of molar mass of gas

Question 38

Graham's law of effusion

Answer 38

The ratio of effusion rates of two different gases is equal to the square root of the division of the molar masses Since rate of effusion is an inverse proportion, it can be said that rate_A times the squre root of A's molar mass is equal to rate_B times the squre root of B's molar mass.

Question 39

Real gases

Answer 39

Do not behave like ideal gases at **high pressure** or **low temperature** Compared to ideal gases: Low pressure: lower ratio of PV/RT High pressure: higher ratio of PV/RT

Question 40

Effect of finite volume of gas particles

Answer 40

Size of gas particles become important at high pressure Because size of particles take up significant portion of total gas volume Hence, volume of the particles must be corrected -- increased

Question 41

Effect of intermolecular forces

Answer 41

At low temperatures, pressure is less than an ideal gas' Because gas atoms spend more time interacting and less colliding, thereby decreasing pressure

Question 42

van der Waals equation

Answer 42

correction for intermolecular forces (P-) × correction for particle volume (V+) = nRT

Question 43

1 atm = ? Pa

Answer 43

1 atm = 101,325 Pa