CHE 101: Gases Lecture Notes

Gases

Gas Substances

Definition: Atmosphere is composed of many gases.
Common gases:
- Usually have diatomic or polyatomic forms.
- Noble gases
- Diatomics
- Ozone (O₃)
- Molecular compounds (e.g. CO, CO₂, N₂O)
Transformation to gas:
- Solids or liquids can be forced to exist as gases called vaporization.
- Example: Water (H₂O) can vaporize into steam.

Physical Characteristics of Gases

Gases automatically take the shape and volume of their container.
- Molecular behavior: Molecules are much farther apart in a gas than in either a solid or liquid.
- There is a lot of empty space (or low density) between molecules.
Highly compressible.
- Gases can be compressed into much smaller volumes.
Always form homogeneous mixtures (solution).

Gases Exert Pressure

Pressure (P): The force exerted by the gas molecules in motion.
Gravity pulls atmospheric gases towards the Earth’s surface.
Typically exert about 14.7 psi (pounds per square inch) at sea level.
Perspective: Car tires usually maintain pressures around 30-35 psi.
Pressure is felt and omnipresent.
Key concept: Why aren't humans crushed by atmospheric pressure?
- The pressure exerted on our bodies is internally countered by our internal pressure equilibrium.
SI Unit: Pascal (Pa).
- Pascal proposed that atmospheric pressure decreases with increasing altitude.
Other common pressure units:
- Atm (atmospheres):
- Pressure at sea level is 1 atm.
- $1 \text{ atm} = 1.013 \times 10^5 \text{ Pa}$ .
- mmHg (millimeter of mercury):
- Derived from barometric pressure measurements.
- $1 \text{ mmHg} = 1 \text{ torr}$ (named after Torricelli, the inventor of the barometer).
- $1 \text{ atm} = 760 \text{ mmHg}$ .
Pressure-Volume Relationship: If air pressure increases, height increases in the context of barometers.

Sample Problem: Convert Pressure

Given: Air pressure in a volleyball is $3.9 \times 10^3 \text{ mmHg}$ .
Question: What is this in atm and torr?
- A. $3.0 \times 10^6$
- B. $5.1$
- C. $3.9 \times 10^3$
- D. $0.74$
Step-by-step solution:
1. Convert to torr: Since $1 \text{ mmHg} = 1 \text{ torr}$ , then $3.9 \times 10^3 \text{ mmHg} = 3.9 \times 10^3 \text{ torr}$ . (Matches option C for torr).
2. Convert to atmospheres (atm): Use the conversion factor $1 \text{ atm} = 760 \text{ mmHg}$ .
  $\text{Pressure in atm} = (3.9 \times 10^3 \text{ mmHg}) \times \left( \frac{1 \text{ atm}}{760 \text{ mmHg}} \right) = 5.13 \text{ atm}$
  (Rounded to two significant figures, this is 5.1 atm, matching option B).

Gas Laws

Four Variables Describing Gases:

Pressure (P)
Volume (V)
Temperature (T)
Amount (n): Number of moles.

1. Boyle’s Law

Relationship: P and V are inversely related.
- If pressure increases, volume decreases.
Mathematical relationship: $P \propto 1/V$ (at constant n and T)
Constant: $PV = \text{constant}$

2. Charles’s Law

Relationship: V and T are directly related.
- If temperature increases, volume also increases.
Visual: Think of an inflating balloon.
Mathematical relationship: $V \propto T$ (at constant n and P)
Constant: Must maintain temperature in Kelvin ( $V/T = \text{constant}$ ).

3. Avogadro’s Law

Relationship: V and n (number of moles) are directly related.
- If the amount of moles increases, volume increases.
Visual: Think of an inflating balloon with more air.
Mathematical relationship: $V \propto n$ (at constant T and P)
Constant: $V/n = \text{constant}$ .

Lecture Summary

Characteristics of gases:
- Constant motion, compressible, and homogeneous.
Pressure definitions & units:
- $1 \text{ atm} = 760 \text{ mmHg} = 760 \text{ torr}$ .
Gas Laws Recap:
- Boyle’s: $P1V1 = P2V2$
- Charles’s: $V1/T1 = V2/T2$
- Avogadro’s: $V1/n1 = V2/n2$

Gas Constant

Gas laws can be combined into one equation to describe ideal gas behavior.
- Boyle’s Law: PV=constant
- Charles's Law: V/T=constant
- Avogadro’s Law: V/n=constant
Ideal Gas Law:
- Combines Boyle’s, Charles’s, and Avogadro's Laws into one equation: $PV = nRT$
- Required parameters for this equation:
- P = pressure
- V = volume
- n = number of moles
- T = temperature in Kelvin ( $TK = TC + 273.15$ ).

Question on Gas Constant (R Value)

Which value of the gas constant should always be used with the Ideal Gas Law?
- Options:
- 0.08206
- 8.314
Answer: The choice of R value depends on the units of Pressure (P) and Volume (V).
- If P is in atmospheres (atm) and V is in liters (L), use $R = 0.08206 \text{ L}\cdot\text{atm}/(\text{mol}\cdot\text{K})$ .
- If P is in Pascals (Pa) and V is in cubic meters ( $m^3$ ), use $R = 8.314 \text{ J}/(\text{mol}\cdot\text{K})$ .
  The question implies multiple choice, but without units specified for P and V, both values are technically correct for different unit systems. However, in most introductory chemistry, $0.08206 \text{ L}\cdot\text{atm}/(\text{mol}\cdot\text{K})$ is commonly used.

Standard Temperature & Pressure (STP)

Definition of STP:
- Standard Temperature = $0°C$ ( $273.15 \text{ K}$ )
- Standard Pressure = $1 \text{ atm}$
Standard Molar Volume:
- Volume of 1.00 mole of gas at STP = $22.4 \text{ L}$ .

Sample: Temperature of Steam

Given: $12.3 \text{g}$ of steam confined to $18.0 \text{ mL}$ at pressure of $1.75 \times 10^6 \text{ torr}$ .
Question: Find the temperature (°C).
Step-by-step solution:
1. Convert given values to Ideal Gas Law units:
- Mass (m) of steam ( $H_2O$ ) = $12.3 \text{ g}$
- Molar mass (Mm) of $H_2O$ = $2(1.008) + 16.00 = 18.016 \text{ g/mol}$
- Number of moles (n) = $m/Mm = 12.3 \text{ g} / 18.016 \text{ g/mol} = 0.6827 \text{ mol}$
- Volume (V) = $18.0 \text{ mL} = 0.0180 \text{ L}$
- Pressure (P) = $1.75 \times 10^6 \text{ torr}$
 - Convert to atm: $P = (1.75 \times 10^6 \text{ torr}) \times (1 \text{ atm} / 760 \text{ torr}) = 2302.6 \text{ atm}$
1. Use the Ideal Gas Law: $PV = nRT$ to solve for T.
- $T = PV / nR$
- Using $R = 0.08206 \text{ L}\cdot\text{atm}/(\text{mol}\cdot\text{K})$
- $T_K = (2302.6 \text{ atm} \times 0.0180 \text{ L}) / (0.6827 \text{ mol} \times 0.08206 \text{ L}\cdot\text{atm}/(\text{mol}\cdot\text{K}))$
- $T_K = 442.2 \text{ K}$
1. Convert Temperature to Celsius:
- $TC = TK - 273.15 = 442.2 - 273.15 = 169.05°C$
- So, the temperature of the steam is approximately $169°C$ .

Group: Helium Gas

Question: How many grams of Helium gas is contained in a $15.0 \text{ L}$ canister with a pressure of $732 \text{ mmHg}$ at $32°C$ ?
Step-by-step solution:
1. Convert given values to Ideal Gas Law units:
- Volume (V) = $15.0 \text{ L}$
- Pressure (P) = $732 \text{ mmHg}$
  - Convert to atm: $P = 732 \text{ mmHg} \times (1 \text{ atm} / 760 \text{ mmHg}) = 0.9632 \text{ atm}$
- Temperature (T) = $32°C$
  - Convert to Kelvin: $T_K = 32 + 273.15 = 305.15 \text{ K}$
- Molar mass (Mm) of Helium (He) = $4.003 \text{ g/mol}$
1. Use Ideal Gas Law ( $PV = nRT$ ) to find the number of moles (n).
- $n = PV / RT$
- $n = (0.9632 \text{ atm} \times 15.0 \text{ L}) / (0.08206 \text{ L}\cdot\text{atm}/(\text{mol}\cdot\text{K}) \times 305.15 \text{ K})$
- $n = 14.448 \text{ L}\cdot\text{atm} / 25.040 \text{ L}\cdot\text{atm/mol} = 0.5770 \text{ mol}$
1. Convert moles to grams:
- Mass (m) = $n \times Mm = 0.5770 \text{ mol} \times 4.003 \text{ g/mol} = 2.31 \text{ g}$
- Thus, the canister contains approximately $2.31 \text{ g}$ of Helium gas.

Combined Gas Law

Definition: Combined Gas Law shows relationships for one specific gas at varying conditions.
Use the Ideal gas equation and solve for one variable in relation to others.
Generic example:
- If a gas has an initial volume of $5.50 \text{ L}$ with a pressure of $0.950 \text{ atm}$ at $25°C$ , what is its volume at STP?
Step-by-step solution for generic example:
1. Identify initial and final conditions:
- Initial: $V1 = 5.50 \text{ L}$ , $P1 = 0.950 \text{ atm}$ , $T_1 = 25°C = 298.15 \text{ K}$
- Final (STP): $V2 = \text{?}$ , $P2 = 1 \text{ atm}$ , $T_2 = 0°C = 273.15 \text{ K}$
1. Use the Combined Gas Law formula (since n is constant):
- $(P1V1)/T1 = (P2V2)/T2$
1. Rearrange to solve for $V_2$ :
- $V2 = (P1V1T2) / (P2T1)$
1. Plug in the values and calculate:
- $V_2 = (0.950 \text{ atm} \times 5.50 \text{ L} \times 273.15 \text{ K}) / (1 \text{ atm} \times 298.15 \text{ K})$
- $V_2 = 1428.1475 \text{ L}\cdot\text{atm}\cdot\text{K} / 298.15 \text{ atm}\cdot\text{K} = 4.79 \text{ L}$
- The volume of the gas at STP is approximately $4.79 \text{ L}$ .
Question: If the pressure in a balloon is maintained at $2.20 \text{ atm}$ , on a day when the temperature is $-15°C$ , what is the volume of the gas on a day when the temperature is $31°C$ ? Which equation is best to use?
Step-by-step solution for question:
1. Determine which gas law applies: The pressure (P) and the amount of gas (n) are constant. The volume (V) and temperature (T) are changing. This indicates Charles's Law.
- Charles's Law: $V1/T1 = V2/T2$
1. Identify initial and final conditions:
- Initial: $P1 = 2.20 \text{ atm}$ (constant), $T1 = -15°C = 258.15 \text{ K}$ , $V_1$ (unknown initial volume, but it cancels out or is implied to exist)
- Final: $P2 = 2.20 \text{ atm}$ (constant), $T2 = 31°C = 304.15 \text{ K}$ , $V_2 = \text{?}$
1. Since we are looking for a relationship between the initial and final volumes, and not an absolute volume, let's assume an initial volume, for instance, $V_1 = 1 \text{ L}$ , or simply express changes proportionally. The question asks for