Ideal Gases

1. The gas laws

• A group of European scientists found three laws between 1662 and 1802, which appeared to explain how gases behaved as their pressure, volume, and temperature changed.
• Since the laws define the mathematical correlations between the three variables, they are entirely empirical and derived from experiments.

⇒ Boyle’s Law:

• Boyle’s law, which describes the relationship between a gas’s pressure and volume, was the first gas law to be found.
• J-shaped tubes filled with sealed glass and mercury were used in the experiments conducted in 1662 by Robert Boyle and his research assistant Robert Hooke.
• Boyle discovered that there was an inverse relationship between the gas’s volume and the pressure it was under.
• Boyle would have produced outcomes such to those displayed in Figure 1.

  
  Figure 1 Boyle’s law graph

• His law is now stated as follows:
  “The pressure of an ideal gas is inversely proportionate to its volume for a fixed mass of the gas at constant temperature”.
  Putting this in mathematical form
• [math]  p \propto \frac{1}{V} \\ p = \text{constant} * \frac{1}{V} \\ p * V = \text{constant} [/math]
• where p represents the pressure (pascal (pa)) (represent on y-axis) applied to a gas of volume V (cubic meter (m³)) (determine on x-axis).
• The same mass of gas at the same temperature, but various pressures and volumes, are used in a more practical form of this equation, where
• [math] p_1 V_1 = p_2 V_2 [/math]

⇒ The pressure-temperature law (Amonton’s law) and absolute zero temperature

• Guillaume Amonton’s attempts to create and construct air thermometers led him to the empirical link between a gas’s temperature and pressure, which he found in 1702.
• Through empirical research, Amonton discovered that, under the assumption that the gas’s mass and volume remained constant, there was a linear connection between the two variables.
• Amonton had difficulty creating precise thermometers, and while his theories were documented, they lacked fundamental quantitative information.
• Figure 4.13 depicts his graph in a more contemporary manner.
• More precisely,
  “The pressure of a constant mass and fixed volume of gas is directly proportional to the gas’s absolute temperature”.
• [math] p \propto T [/math]
• Or
• [math] p = \text{constant} (T) \\  \frac{p}{T} = \text{constant} [/math]
• 
  Figure 2 Amonton pressure-temperature gas law
• A third, more practical version of this connection is utilized to compare the same gas at various temperatures and pressures. This can be expressed as;
• [math] \frac{p_1}{T_1} = \frac{p_2}{T_2} [/math]
• At the time of his tests, Amonton understood that there would come a temperature at which a gas’s pressure would zero out if he continued to extrapolate his results through ever-lower temperatures.
• The molecules would cease moving at this temperature, making it impossible for them to provide pressure by colliding with another object.
• Amonton determined the temperature, which became known as absolute zero, to be -240°C.
• This was a rather amazing accomplishment given the thermometer technology at the time (air thermometers), since he was only around 35°C off.

⇒ Charles law

• A study by French scientist Joseph Gay-Lussac from 1802, demonstrating the experimental relationship between a gas’s volume and temperature, was published. Gay-Lussac gave the rule its name in honor of his friend, the balloonist Jacques Charles, who, after observing balloon activity, created an unpublished version of the law.

  
  Figure 3 Charles law graph

• The graph in Figure 3 provides an illustration of the empirical law.
  “At constant pressure the volume of a fixed mass of an ideal gas is directly proportional to its absolute temperature”
• Expressing this mathematically:
• [math] V \propto T \\  V = \text{constant} (T) \\  \frac{V}{T} = \text{constant} [/math]
• Once more, using the Kelvin scale (T/KT/°C + 273.15), the temperature T is an absolute value.
• Like the other gas laws, this connection also has another practical version that is
• [math]  \frac{V_1}{T_1} = \frac{V_2}{T_2} [/math]
• The units of measurement for pressure in combined gas law problems are often pascals (Pa) or kilopascals (kPa), where 1 Pa is equal to 1 N acting over 1 m.
• Additionally, the unit of measurement for pressure is atmospheres (atm), where a standard atmosphere is the average atmospheric pressure at sea level and is specified as 101 325 Pa.
• While cm³, litres, and millilitres are also often used, m³ is the standard unit of measurement for gas volume.
• Temperatures can be expressed in degrees Celsius or kelvin. It is necessary to convert the latter to kelvin.

Standard temperature-pressure (STP):

• Standard temperature and pressure (STP) are defined as:
  – Temperature: 20°C (293.15 K)
  – Pressure: 1 atm (1013 mbar)
• These conditions are used as a reference point for many scientific and engineering applications, such as:
  1. Gas laws and equations
  2. Chemical reactions and kinetics
  3. Thermodynamic calculations
  4. Material properties and testing
• Using STP conditions helps to:
  1. Simplify comparisons and calculations
  2. Ensure consistency and accuracy
  3. Provide a common reference point for different experiments and data
• Some key aspects of STP:
  1. Temperature: 20°C is a convenient and easily achievable temperature, close to room temperature.
  2. Pressure: 1 atm is the average atmospheric pressure at sea level, making it a logical choice.
  3. Universal applicability: STP conditions are widely accepted and used across various fields and industries.

2. Absolute zero:

• At the time of his tests, Amonton understood that there would come a temperature at which a gas’s pressure would zero out if he continued to extrapolate his results through ever-lower temperatures.
• The molecules would cease moving at this temperature, making it impossible for them to provide pressure by colliding with another object.
• Amonton determined the temperature, which became known as absolute zero, to be -240°C.
• This was a rather amazing accomplishment given the thermometer technology at the time (air thermometers), since he was only around 35°C off.
  “Absolute zero is the lowest possible temperature, defined as 0 Kelvin (K), -273.15 degrees Celsius (°C), or -459.67 degrees Fahrenheit (°F)”.
• It is the temperature at which all matter would theoretically have zero entropy and zero kinetic energy.
• At absolute zero:

  
  Figure 4 absolute zero graph

  1. All molecular motion ceases (classical physics perspective)
  2. Quantum fluctuations still occur (quantum mechanics perspective)Entropy is minimized
  3. Entropy is minimized
  4. Perfect crystal structure (no defects or randomness)
• Reaching absolute zero is impossible, as it would require the removal of all entropy, which violates the second law of thermodynamics.
• However, scientists have achieved temperatures very close to absolute zero using advanced techniques like laser cooling and magnetic trapping.
• Some interesting facts about absolute zero:
  1. Absolute zero is a fundamental limit: It’s the lowest possible temperature, and any attempt to go below it would violate the laws of physics.
  2. Unattainable: As mentioned earlier, it’s impossible to reach absolute zero, but we can get very close!
  3. Quantum weirdness: Even at absolute zero, quantum fluctuations occur, which is mind-bending to think about!
  4. Perfect crystal structure: At absolute zero, materials would have a perfect crystal structure, which is fascinating from a materials science perspective.

⇒Combining the gas laws

• A combined gas law, which links a gas’s pressure, volume, and temperature, may be mathematically represented as follows.
• This combines the three gas laws: Boyle’s law, Amonton’s law, and Charles law.
• [math] \frac{pV}{T} = \text{constant} [/math]
• A more useful form of this equation can be written as
• [math] \frac{p_1 V_1}{T_1} = \frac{p_2 V_2}{T_2} [/math]
• where [math]P_2 , V_2 , and \,T_2[/math] represent the final pressure, volume, and temperature of the same gas after a modification has been made, and [math] P_1 , V_1 , and \,T_1 [/math] represent the initial pressure, volume, and temperature of a gas.
• Example:
• A balloon’s volume changing
  A party balloon has a capacity of 2400 cm³ when it is at ambient temperature (15°C) and atmospheric pressure (1.0 atm) at sea level. The balloon is flown up a mountain where the air pressure has decreased to 0.80 atm and the temperature is 1.4°C. Determine the balloon’s new volume at the summit of the mountain.
  Solution:
  Use
• [math] \frac{p_1 V_1}{T_1} = \frac{p_2 V_2}{T_2} [/math]
• Substituting numbers:
• [math] \frac{(1)(2400)}{(15 + 273.15) \, \text{K}} = \frac{(0.80) V_2}{(1.4 + 273.15) \, \text{K}} [/math]

• Rearranging gives:

• [math] V_2 = \frac{(1)(2400)(1.4 + 273.15)}{(15 + 273.15)(0.80)} \\ V_2 = 2858.4 \, \text{cm}^3 [/math]

3. Avogadro’s law, the ideal gas equation and moles:

• Amadeo Avogadro postulated and published a fourth experimental gas law in 1811.
• According to this law, “there were the same number of molecules in equal quantities of gases at the same temperature and pressure”.
• In terms of math, this indicates that:
• [math] V \propto n \\ V = \text{constant} * n
  \\ \frac{V}{n} = \text{constant} [/math]
• where n is the gas’s molecular weight. Similar to the other gas laws, this is best expressed as follows:
• [math] \frac{V_1}{n_1} = \frac{V_2}{n_2} [/math]
• The ideal gas equation was created in 1834 by French scientist Emile Clapeyron by combining all four gas laws:
• [math] pV=nRT \qquad (1) [/math]
• where [math] R = 8.31Jmol^{-1}K^{-1}[/math] is the constant, which is now referred to as the molar gas constant.
• The ideal gas equation is a very strong formula that accurately describes the behavior of gases, especially at high temperatures and very low pressures.
  It performs less effectively at circumstances near liquefaction because the gases behave substantially differently from ideal gases.
• Jean Baptiste Perrin proposed the Avogadro constant, [math] N_A[/math], in 1909, A to indicate the quantity of particles (typically molecules or atoms) in one mole (1 mol) of a material.
• It was given the name Amadeo Avogadro by Perrin.
• Perrin’s efforts to precisely estimate the Avogadro constant ultimately led him to be awarded the Nobel Prize in Physics.
• According to the AQA Physics Datasheet, the Avogadro constant is now specified as
  [math] 6.02214129 * 10 ^{23} mol^{-1}[/math]  (rounded to [math] 6.02 * 10^{23} mol^{-1}[/math]).
  The Boltzmann constant, k, is another basic constant in physics that is obtained by dividing the molar gas constant, R, by the Avogadro constant, [math] N_A[/math].
• [math] k = \frac{R}{N_A} [/math]
• Putting values
• [math] k = \frac{R}{N_A} = \frac{8.31}{6.02 \times 10^{23}} = 1.38 \times 10^{-23} \, \text{J K}^{-1} [/math]
• The basic constant that connects the microscopic behavior of particles in a gas to the macroscopic measurements of pressure, volume, and temperature is known as the Boltzmann constant.
• It plays a crucial role in the ideal gas model.
• The constant allows ideal gases to be objectively measured by the gas laws, allowing the microscopic model to predict their behavior on a macroscopic scale.
• The above equation can be rearranged to give:
• [math] R = kN_A [/math]
• which can be substituted into the ideal gas equation (1),giving
• [math] pV=nN_A kT [/math]
• But [math] nN_A [/math] is the number of particles in the gas and is given the symbol (N) so:
• [math] pV=NkT [/math]

Example:
⇒ A cylinder of propane gas
The capacity of a 0.14 m³ propane gas cylinder is filled to a pressure of [math]2.0 \times 10^6[/math] Pa above atmospheric pressure at 300K.
Solution:

1. Determine how many moles of propane gas are contained in the cylinder.
Solution:
[math] pV = nRT [/math] is the ideal gas equation.

[math] n = \frac{pV}{RT} \\
n = \frac{(2 \times 10^6 + 1 \times 10^5)(0.14)}{(8.31)(300)} = 118 \, \text{mol} [/math]

2. If the molar mass of propane is [math] 44.1 g.mol^{-1}[/math], find the mass of gas within the cylinder.
Solution:
Since there are 118 mol of gas and [math]44.1 g.mol^{-1}[/math] of propane in the cylinder, the mass of propane within is

[math] mass of propane=44.1*118 \\
m=5204g\\
m=5.2kg [/math]

3-Calculate the mass of one molecule of propane
Solution:
The mass of one molecule of propane is

[math] m = \frac{\text{molar mass}}{N_A} \\
m = \frac{44.1 \times 10^{-3}}{6.02 \times 10^{23}} \\
m = 7.3 \times 10^{-26} \, \text{kg} [/math]

4.Work done in thermodynamics (gas process):

• “The energy transferred from one thermodynamic system into another thermodynamic system”.
• [math] W = p \Delta V [/math]
•  Relating to the work done by a gas during a process.
  – Work done (W): The energy transferred from or to the system due to a change in volume.
  – p (pressure): The force exerted per unit area on the gas.
  – ∆V (change in volume): The difference between the initial and final volumes of the gas.
• The equation states that the work done by a gas is equal to the pressure multiplied by the change in volume.
• This means that if the volume of a gas increases, the gas does work on its surroundings, and if the volume decreases, work is done on the gas.
• Some aspects:
  – Units: Work done is typically measured in Joules (J), pressure in Pascals (Pa), and volume in cubic meters (m³).
  – Sign convention: Positive work done indicates work done by the system (expansion), while negative work done indicates work done on the system (compression).
• Applications: This equation is widely used in various fields, such as:
  – Engine cycles (e.g., Carnot cycle, Otto cycle)
  – Thermodynamic processes (e.g., isothermal, adiabatic)
  – Gas compression and expansion

5. Molar mass and molecular mass:

⇒ Molar mass:

• “Molar mass is the mass of one mole of a substance, expressed in grams per mole (g/mol)”.
• It represents the total mass of all the atoms in a molecule, taking into account the atomic masses of each element.
• Molar mass is calculated by summing the atomic masses of all the atoms in a molecule. For example:
  – Water (H2O): 2(1.00794 u) + 15.9994 u = 18.0153 g/mol
  – Carbon dioxide (CO2): 12.0107 u + 2(15.9994 u) = 44.0095 g/mol
• Molar mass is used in various applications, such as:
  1. Chemical reactions: To calculate the amounts of reactants and products.
  2. Gas laws: To determine the behavior of gases.
  3. Concentration calculations: To prepare solutions and calculate their concentrations.
  4. Material properties: To determine the properties of materials, such as density and volume.
• Some key points to remember:
  – Molar mass is a property of a substance, not a mixture.
  – It is expressed in units of g/mol.
  – It is used to convert between moles and grams.

⇒ Molecular mass:

• “Molecular mass, also known as molecular weight, is the sum of the atomic masses of all the atoms in a molecule”.
• It’s a measure of the total mass of a single molecule, taking into account the atomic masses of each element.
  1. Atomic masses: The mass of each atom in the molecule, usually expressed in units of u (unified atomic mass units) or amu (atomic mass units).
  2. Molecular formula: The formula that shows the number of atoms of each element in the molecule.
  3. Molecular mass calculation: Add up the atomic masses of all the atoms in the molecule, using the molecular formula.
• For example, let’s calculate the molecular mass of glucose (C₆H₁₂O₆):
• Atomic masses:
  – Carbon (C): 12.0107 u
  – Hydrogen (H): 1.00794 u
  – Oxygen (O): 15.9994 u
• Molecular formula: C₆H₁₂O₆
• Molecular mass calculation:
  – 6 Carbon atoms: 6 x 12.0107 u = 72.0642 u
  – 12 Hydrogen atoms: 12 x 1.00794 u = 12.09528 u
  – 6 Oxygen atoms: 6 x 15.9994 u = 95.9964 u
  – Total molecular mass: 72.0642 u + 12.09528 u + 95.9964 u = 180.15588 u
• Rounded to the nearest whole number, the molecular mass of glucose is 180 g/mol.
• Utilising the mass of particles and counting sizable clusters of them with an electronic balance is a more accurate method of calculating the amount of matter in a substance.
• The number of particles present can be determined by measuring the masses of several of them if the mass of one particle is known.

• As a result, we define two quantities: the mass of a single substance molecule, known as the molecular mass (m), and the mass of a single mole ([math]N_A[/math]) of molecules of the material, known as the molar mass ([math]M_m[/math]). The relationship between these two quantities is as follows:

• [math] M_m=N_A m [/math]
• When the mass of a known gas, Mg, is measured, the number of moles, n, is obtained by dividing it by the molar mass; the number of molecules, N, is obtained by dividing it by the molecular mass, m:
• [math] n = \frac{M_g}{M_m} \\
  n = \frac{M_g}{N_A \cdot m} \\
  n N_A = N = \frac{M_g}{m} [/math]
• Substituting both of them into the ideal gas equation enables macroscopically measurement of all quantities:
• [math] pV = \frac{M_g}{M_m} \cdot RT [/math]
• And
• [math] pV = \frac{M_g}{m} \cdot kT [/math]