Ideal Gases

Ideal Gases

• Gases are assumed to be ideal under varying conditions of temperature and pressure.
• An ideal gas is a hypothetical gas whose molecules have no intermolecular forces and occupy no volume.
• Real gases deviate from ideal behavior at high pressures, low volumes, and low temperatures.
• Many compressed real gases behave close to ideal.

Ideal Gas Law

• First stated in 1834 by Benoit Paul Emil Clapeyron.

• Boyle's law, Charles' law, and Dalton's law were established before the ideal gas law.

• The ideal gas law shows the relationship among four variables that define a gas.

  $PV = nRT$

  • P = pressure
  • V = volume
  • n = number of moles
  • T = temperature
  • R = the ideal gas constant

• $R = 8.21 \times 10^{-2} L \cdot ATM / (mol \cdot K)$

• R can be expressed in other units.

• $R = 8.314 J / (K \cdot mol)$

• The ideal gas law is used to determine the missing term when given all of the others.

• It can be used to calculate the change in a term while holding two of the others constant.

• It is most commonly used to solve the volume or pressure at any given temperature and number of moles.

Example

• What volume would $12$ grams of helium occupy at $27$ degrees Celsius and a pressure of $380$ mmHg?

Solution

• The ideal gas law can be used, but first all of the variables must be converted to units that will correspond to the expression of the gas constant as $8.21 \times 10^{-2} L \cdot ATM / (mol \cdot K)$.
• $P = 380 mmHg \times (1 ATM / 760 mmHg) = 0.5 ATM$
• $T = 27 \degree C + 273 = 300 K$
• $n = 12 g He \times (1 mol / 4.0 g) = 3 mol He$
• $PV = nRT$
• $0.5 ATM \times V = 3 mol He \times 0.0821 L \cdot ATM / (mol \cdot K) \times 300 K$
• $0.5V \approx 3 \times 8 \times 3$
• $V = 144 L$
• The actual value equals $148 L$.

Density

• Density is defined as the ratio of the mass per unit volume of a substance.

• The densities of gases are usually expressed in units of grams per liter.

• The ideal gas law contains variables for volume and number of moles.

• The law can be rearranged to calculate the density of any gas.

  $PV = nRT$

  $n = m / M$

  • m = mass
  • M = molar mass

  $PV = (m/M)RT$

  $p = m/V = (P \times M) / (RT)$

• A mole of an ideal gas at STP occupies $22.4 L$.

Combined Gas Law

• Can be used to relate changes in temperature, volume, and pressure of a gas.

  $(P<em>1 \times V</em>1) / T<em>1 = (P</em>2 \times V<em>2) / T</em>2$

  • The subscripts one and two refer to the two states of the gas at STP and at the conditions of actual temperature and pressure.

• This equation assumes the number of moles stays constant.

• To calculate a change in volume, the equation is rearranged as follows:

  $V<em>2 = V</em>1 \times (P<em>1 / P</em>2) \times (T<em>2 / T</em>1)$

  • $V_2$ is then used to find density of the gas under nonstandard conditions.

    $p = m / V_2$

• Doubling the temperature of a gas would result in doubling its volume.

• Doubling the pressure of a gas would result in halving the volume.

• Doubling both the temperature and pressure at the same time results in a final volume that is equal to the original volume.

Example

• What is the density of $CO_2$ gas at $2$ ATM and $273$ degrees C?

Solution

• At STP, a mole of gas occupies $22.4 L$.

• Because the increase in pressure to $2$ ATM decreases volume proportionally, $22.4 L$ must be multiplied by $1 ATM / 2 ATM = 0.5$.

• Because the increased temperature increases volume proportionally, the temperature factor will be $546 K / 273 K = 2$.

  $V_2 = (22.4 L / mol) \times (1 ATM / 2 ATM) \times (546 K / 273 K) = 22.4 L / mol$

  $p = (44 g / mol) / (22.4 L / mol) \approx 2 g / L$

  • The actual value equals $1.96 g / L$.

Molar Mass

• Sometimes the identity of a gas is unknown.

• The molar mass can be determined in order to identify the gas.

• Using the equation for density derived from the ideal gas law, we can calculate the molar mass of a gas experimentally.

• The pressure and temperature of a gas contained in a bulb of a given volume are measured and the mass of the bulb with the sample is measured.

• Then the bulb is evacuated, the gas is removed, and the mass of the empty bulb is determined.

• The mass of the bulb with the sample minus the mass of the evacuated bulb gives the mass of the sample.

• Finally, the density of the sample is determined by dividing the mass of the sample by the volume of the bulb.

• This gives the density at the given temperature and pressure.

  $V<em>2 = V</em>1 \times (P<em>1 / P</em>2) \times (T<em>2 / T</em>1)$

  • Substitute in $273 K$ for $T<em>2$ and $1 ATM$ for $P</em>2$.

• The ratio of the sample mass divided by $V_2$ gives the density of the gas at STP.

• The molar mass can then be calculated as the product of the gas's density at STP and the STP volume of one mole of gas, $22.4 L / mol$.

  $M = p_{STP} \times 22.4 L / mol$

Example

• What is the molar mass of a $22.4 L$ sample of gas that has a mass of $225 g$ at a temperature of $273$ degrees C and a pressure of $10 ATM$?

Solution

• Determine how the current conditions compare to STP and use this to set up proportional relationship.

  $(P<em>1 \times V</em>1) / T<em>1 = (P</em>2 \times V<em>2) / T</em>2$

  $V<em>{STP} = V</em>1 \times (P<em>1 / P</em>{STP}) \times (T<em>{STP} / T</em>1)$

  $V_{STP} = 22.4 L \times (10 ATM / 1 ATM) \times (273 K / 546 K) = 22.4 \times 10 \times (1/2) = 112 L$

  $225 g / 112 L \approx 2 g / L at STP$

  $M = 2 g / L \times 22.4 L / mol \approx 44.8 g / mol$

Special Cases

• Even though the following laws were developed before the ideal gas law, it is conceptually helpful to think of them as special cases of the more general ideal gas law.

Avogadro's Principle

• States that all gases at a constant temperature and pressure occupy volumes that are directly proportional to the number of moles of gas present.

  $n / V = k$

  $n<em>1 / V</em>1 = n<em>2 / V</em>2$

  • k is a constant
  • $n<em>1$ and $n</em>2$ are the number of moles of gas $1$ and gas $2$ respectively
  • $V<em>1$ and $V</em>2$ are volumes of the gases respectively

• As the number of moles of gas increases, the number the volume increases in direct proportion.

Example

• A $2.0 L$ sample at a $100$ degrees C and $20 ATM$ contains $5$ moles of gas. If an additional $25$ moles of gas at the same pressure and temperature are added, what is the final volume of the gas?

Solution

• If pressure and temperature are held constant, the ideal gas law reduces to Avogadro's principle.

  $n<em>1 / V</em>1 = n<em>2 / V</em>2$

  $5 mol / 2.0 L = (5 mol + 25 mol) / V_2$

  $V_2 = (30 mol \times 2.0 L) / 5 mol = 12 L$

Boyle's Law

• For a given gaseous sample held at constant temperature (isothermal conditions), the volume of the gas is inversely proportional to its pressure.

  $PV = k$

  $P<em>1 \times V</em>1 = P<em>2 \times V</em>2$

  • k is a constant
  • The subscripts one and two represent two different sets of pressure and volume conditions.

• Boyle's law is simply the special case of the ideal gas law in which n and t are constant.

Example

• What would be the volume of a one liter sample of helium if its pressure is changed from $12 ATM$ to $4 ATM$ under isothermal conditions?

Solution

• If the number of moles of gas and temperature are held constant, the ideal gas law reduces to Boyle's law.

  $P<em>1 \times V</em>1 = P<em>2 \times V</em>2$

  $12 ATM \times 1 L = 4 ATM \times V_2$

  $V_2 = (12 ATM \times 1 L) / 4 ATM = 3 L$

Charles’s Law

• At constant pressure, the volume of a gas is proportional to its absolute temperature expressed in Kelvins.

  $V / T = k$

  $V<em>1 / T</em>1 = V<em>2 / T</em>2$

  • k is a proportionality constant
  • The subscripts one and two represent two different sets of temperature and volume conditions.

• Charles' law is another special case of the ideal gas law in which n and p are constant.

Example

• If the temperature of $2 L$ of a gas at constant pressure is, changed from $290 K$ to $580 K$, what would be its final volume?

Solution

• If the number of moles of gas and pressure are held constant, the ideal gas law reduces to Charles law.

  $V<em>1 / T</em>1 = V<em>2 / T</em>2$

  $2 L / 290 K = V_2 / 580 K$

  $V_2 = (2 L \times 580 K) / 290 K = 4 L$

Gay-Lussac's Law

• Relates pressure to temperature, n and v are constant.

  $P / T = k$

  $P<em>1 / T</em>1 = P<em>2 / T</em>2$

  • k is proportionality constant
  • The subscripts one and two represent two different sets of temperature and pressure conditions.

• An increase in temperature will increase the pressure in direct proportion.

Example

• If the pressure of a sample of gas with a temperature of $300 K$ changes from $2 ATM$ to $5 ATM$ during heating, what would be the final temperature if volume is held constant?

Solution

• If the number of moles of gas and volume are held constant, the ideal gas law reduces to Gay-Lussac's law.

  $P<em>1 / T</em>1 = P<em>2 / T</em>2$

  $2 ATM / 300 K = 5 ATM / T_2$

  $T_2 = (5 ATM \times 300 K) / 2 ATM = 750 K$

Combined Gas Law

• Relates pressure and volume (Boyle's law) in the numerator.
• Relates the variation in temperature to both volume (Charles' law) and pressure (Gay-Lussac's law) simultaneously.

Dalton's Law of Partial Pressures

• When two or more gases that do not chemically interact are found in one vessel, each gas will behave independently of the others.

• The pressure exerted by each gas in the mixture will be equal to the pressure that the gas would exert if it were the only one in the container.

• The pressure exerted by each individual gas is called the partial pressure of that gas.

• The total pressure of a gaseous mixture is equal to the sum of the partial pressures of components.

  $P<em>t = P</em>A + P<em>B + P</em>C + …$

  • $P_t$ is the total pressure in the container
  • $P<em>A$, $P</em>B$, $P_C$ are the partial pressures of gases A, B, and C respectively

• The partial pressure of a gas is related to its mole fraction and can be determined using the following equation.

  $P<em>A = X</em>A \times P_t$

  $X_A = moles \ of \ gas \ A \ over \ total \ moles \ of \ gas$

Example

• A vessel contains $0.75 mol$ of nitrogen, $0.2 mol$ of hydrogen, and $0.05 mol$ of fluorine at a total pressure of $2.5 atm$. What is the partial pressure of each gas?

Solution

• First, calculate the mole fraction of each gas.

  $X<em>{N</em>2} = 0.75 mol / 1.00 mol = 0.75$

  $X<em>{H</em>2} = 0.2 mol / 1.00 mol = 0.2$

  $X<em>{F</em>2} = 0.05 mol / 1.00 mol = 0.05$

• Then calculate the partial pressure.

  $P<em>A = X</em>A \times P_t$

  $P<em>{N</em>2} = 0.75 \times 2.5 atm = 1.875 atm$

  $P<em>{H</em>2} = 0.2 \times 2.5 atm = 0.5 atm$

  $P<em>{F</em>2} = 0.05 \times 2.5 atm = 0.125 atm$

Henry's Law

• At various applied pressures, the concentration of a gas in a liquid increased or decreased.

• This was a characteristic of a gas's vapor pressure.

• Vapor pressure is the pressure exerted by evaporated particles above the surface of a liquid.

  Evaporation is a dynamic process that requires the molecules at the surface of a liquid to gain enough energy to escape into the gas phase.

  Vapor pressure from the evaporated molecules forces some of the gas back into the liquid phase, and equilibrium is reached between evaporation and condensation.

• Mathematically, this is expressed as:

  $a = k<em>H \times P</em>A$

  $a<em>1 / P</em>1 = a<em>2 / P</em>2 = k_H$

  • a is the concentration of a in solution
  • $k_H$ is Henry's constant
  • $P_A$ is the partial pressure of a

• The value of Henry's constant depends on the identity of the gas.

• According to this relationship, solubility and pressure are directly related.

• If the partial pressure of a particular gas is elevated, such as when given hyperbaric oxygen, the amount of that gas dissolved in the blood is also elevated.

Example

• If $4 \times 10^{-4}$ moles of gas are dissolved in $2 L$ of solution under an ambient pressure of $2 ATM$, what will be the molar concentration of the gas under $10 ATM$?

Solution

• Start by determining the initial concentration of the gas in solution.

  $a_1 = (4 \times 10^{-4} mol) / 2 L = 2 \times 10^{-4} M$

• Next, utilize the direct relationship between solubility and pressure according to Henry's law.

  $a<em>1 / P</em>1 = a<em>2 / P</em>2$

  $(2 \times 10^{-4} M) / 2 ATM = a_2 / 10 ATM$

  $a_2 = (2 \times 10^{-4} M \times 10 ATM) / 2 ATM = 10^{-3} M$