Gas Laws and Pressure Summary

7.2 Gases and Pressure

A. Properties of Gases

The Kinetic-Molecular Theory of Gases:

• Gases are composed of small and widely separated particles (molecules or atoms).
• Particles of a gas behave independently of one another.
• Each particle of a gas is in rapid, straight-line motion until it collides with another molecule or with its container.
• The pressure of a gas arises from the sum of the collisions of the particles with the walls of the container.
• The average kinetic energy of gas particles depends only on the absolute temperature: $KE<em>{ave} \propto T</em>{Kelvin}$

7.2 Gases and Pressure

B. Gas Pressure

Definition of Pressure:

• When gas particles collide with the walls of a container, they exert pressure.
• Pressure (P) is the force (F) exerted per unit area (A): $P = \frac{Force}{Area} = \frac{F}{A}$

Units of Pressure:

• 1 atmosphere (atm) = 760 mmHg = 760 torr = 14.7 psi = 101325 Pa (Pascals, where Pa = N/m²) = 101.325 kPa

Activity: Pressure Conversions

• Problem: The pressure of a gas is $8.25 \times 10^4 Pa$. What is this pressure expressed in units of atm and torr?
• Given:
  • 1 atm = 760 mmHg
  • 1 atm = 760 torr
  • 1 atm = 101,325 Pa

Activity Solutions: Pressure Conversions

• Conversion of $8.25 \times 10^4 Pa$ to atm:
  $8.25 \times 10^4 Pa \times \frac{1 atm}{101,325 Pa} = 8.14 \times 10^{-1} atm$

• Conversion of $8.14 \times 10^{-1} atm$ to torr:
  $8.14 \times 10^{-1} atm \times \frac{760 torr}{1 atm} = 6.19 \times 10^2 torr$

7.3 Gas Laws

A. Boyle’s Law

Definition:

• For a fixed amount of gas at constant temperature, the pressure and volume of the gas are inversely related.
• If one quantity increases, the other decreases.
• The product of the two quantities is a constant, k: $Pressure \times Volume = constant$
• $P \times V = k$

Explanation:

• If the volume of a cylinder of gas is halved, the pressure of the gas inside the cylinder doubles.
• This behavior can be explained by the equation: $P<em>1V</em>1 = P<em>2V</em>2$
  • $P<em>1, V</em>1$ = initial conditions
  • $P<em>2, V</em>2$ = new conditions

How to Use Boyle’s Law to Calculate a New Gas Volume or Pressure:

• Example: If a 4.0-L container of helium gas has a pressure of 10.0 atm, what pressure does the gas exert if the volume is increased to 6.0 L at constant temperature?
  • Step 1: Identify the known quantities and the desired quantity.
    • $P_1 = 10.0 atm$
    • $V_1 = 4.0 L$
    • $V_2 = 6.0 L$
    • $P_2 = ?$
  • Step 2: Write the equation and rearrange it to isolate the desired quantity on one side.
    • $P<em>1V</em>1 = P<em>2V</em>2$
    • Solve for $P<em>2$ by dividing both sides by $V</em>2$: $P<em>2 = \frac{P</em>1V<em>1}{V</em>2}$
  • Step 3: Solve the problem.
    • $P_2 = \frac{(10.0 atm)(4.0 L)}{6.0 L}$
    • $P_2 = 6.67 atm$

B. Charles’s Law

Definition

• For a fixed amount of gas at constant pressure, the volume of the gas is proportional to its Kelvin temperature.
• If one quantity increases, the other increases as well.
• Dividing volume by temperature is a constant, k: $\frac{Volume}{Temperature} = constant$
• $\frac{V}{T} = k$

Explanation:

• If the temperature of the cylinder is doubled, the volume of the gas inside the cylinder doubles.
• This behavior can be explained by the equation: $\frac{V<em>1}{T</em>1} = \frac{V<em>2}{T</em>2}$
  • $V<em>1, T</em>1$ = initial conditions
  • $V<em>2, T</em>2$ = new conditions

C. Gay–Lussac’s Law

Definition:

• For a fixed amount of gas at constant volume, the pressure of a gas is proportional to its Kelvin temperature.
• If one quantity increases, the other increases as well.
• Dividing pressure by temperature is a constant, k: $\frac{Pressure}{Temperature} = constant$
• $\frac{P}{T} = k$

Explanation:

• Increasing the temperature increases the kinetic energy of the gas particles, causing the pressure exerted by the particles to increase.
• This behavior can be explained by the equation: $\frac{P<em>1}{T</em>1} = \frac{P<em>2}{T</em>2}$
  • $P<em>1, T</em>1$ = initial conditions
  • $P<em>2, T</em>2$ = new conditions

D. The Combined Gas Law

Definition:

• All three gas laws can be combined into one equation: $\frac{P<em>1V</em>1}{T<em>1} = \frac{P</em>2V<em>2}{T</em>2}$
• This equation is used for determining the effect of changing two factors (for example, P and T) on the third factor (V).

Summary of the Gas Laws Table:

7.4 Avogadro’s Law

Definition:

• When the pressure and temperature are held constant, the volume of a gas is proportional to the number of moles present.
• If one quantity increases, the other increases as well.
• Dividing the volume by the number of moles is a constant, k: $\frac{Volume}{Number \ of \ moles} = constant$
• $\frac{V}{n} = k$

Avogadro’s Law Equation:

• $\frac{V<em>1}{n</em>1} = \frac{V<em>2}{n</em>2}$
  • $V<em>1, n</em>1$ = initial conditions
  • $V<em>2, n</em>2$ = new conditions

Standard Temperature and Pressure (STP):

• Often, amounts of gas are compared at a set of standard conditions of temperature and pressure, abbreviated as STP.
• STP conditions are:
  • 1 atm (760 mm Hg) for pressure
  • 273 K (0°C) for temperature
• At STP, 1.00 mole of any gas has a volume of 22.4 L. 22.4 L is called the standard molar volume.

How to Convert Moles of Gas to Volume at STP:

• Example: How many moles are contained in 2.0 L of N2 at standard temperature and pressure (STP)?
  • Step 1: Identify the known quantities and the desired quantity.
    • 2. 0 L of N2 (original quantity)
    • ? moles of N2 (desired quantity)
  • Step 2: Write out the conversion factors.
    • $\frac{22.4 L}{1 mol} or \frac{1 mol}{22.4 L}$
    • Choose the right side to cancel L.
  • Step 3: Set up and solve the problem.
    • $2.0 L \times \frac{1 mol}{22.4 L} = 0.089 mol \ N_2$
    • Liters cancel.

Activity: The Combined Gas Law Sample

• A sample of hydrogen gas occupies 1.25 L at 80.0°C and 2.75 atm. What volume will it occupy at 185°C and 5.00 atm?

Activity Solution: The Combined Gas Law

• First, convert the temperatures into kelvins:
  • $T_1 = 80.0°C + 273.15 K = 353.2 K$
  • $T_2 = 185°C + 273.15 K = 458 K$
• Given:
  • $V_1 = 1.25 L$
  • $V_2 = ?$
  • $T_1 = 353.2 K$
  • $T_2 = 458 K$
  • $P_1 = 2.75 atm$
  • $P_2 = 5.00 atm$

Activity Solution: The Combined Gas Law

Combined Gas Law:

• $\frac{P<em>1V</em>1}{T<em>1} = \frac{P</em>2V<em>2}{T</em>2}$
• Solving for $V_2$:
  • $V<em>2 = \frac{P</em>1V<em>1T</em>2}{T<em>1P</em>2}$
  • Getting numerical value:
    • $V_2 = \frac{2.75 atm \times 1.25 L \times 458 K}{353.2 K \times 5.00 atm} = 0.891 L$

7.5 The Ideal Gas Law

Definition:

• All four properties of gases (P, V, n, and T) can be combined into a single equation called the ideal gas law.
• $PV = nRT$

Universal Gas Constant (R):

• $R = 0.0821 \frac{L \cdot atm}{mol \cdot K}$
  • $R$ can be derived from the ideal gas law:
    • $R = \frac{PV}{nT}$

How to carry out Calculation with the Ideal Gas Law

• Example: How many moles of gas are contained in a typical human breath that takes in 0.50 L of air at 1.0 atm pressure and 37°C?
  • Step 1: Identify the known quantities and the desired quantity.
    • $P = 1.0 atm$
    • $V = 0.50 L$
    • $T = 37 °C$
    • $n = ? mol$
  • Step 2: Convert all values to proper units and choose the value of R that contains these units.
    • Temperature is given in °C, but must be in K: $K = °C + 273$
    • $K = 37°C + 273= 310 K$
    • The pressure is given in atm, so use the following R value:
      • $R = 0.0821 \frac{L \cdot atm}{mol \cdot K}$
  • Step 3: Write the equation and rearrange it to isolate the desired quantity on one side.
    • $PV = nRT$
    • Solve for n by dividing both sides by RT: $n = \frac{PV}{RT}$
  • Step 4: Solve the problem.
    • $n = \frac{(1.0 atm)(0.50 L)}{(0.0821 \frac{L \cdot atm}{mol \cdot K})(310 K)} = 0.020 mol$

Example using Ideal Gas Law at STP

• How many moles are contained in 2.0L of $N_2$ at standard temperature and pressure (STP)?
• Calculate the volume of 1.00 mol of an ideal gas at standard temperature and pressure (STP).

Consider the reaction:

• $Zn(s) + 2HCl(aq) \rightarrow ZnCl<em>2(aq) + H</em>2(g)$. What volume of $H_2(g)$ at STP can be generated when 134 g of zinc reacts? Answer: 45.9 L
• Consider the reaction: $2H<em>2O</em>2(l) \rightarrow 2H<em>2O(l) + O</em>2(g)$. If 3.0 mol of $H<em>2O</em>2$ decompose, what volume of $O_2(g)$ is produced at 1.0 atm and 23°C? Answer: 36 L

7.6 Dalton’s Law and Partial Pressures

Definition:

• Dalton’s law: The total pressure $(P_{total})$ of a gas mixture is the sum of the partial pressures of its component gases.
• For a mixture of three gases A, B, and C:
  • $P<em>{total} = P</em>A + P<em>B + P</em>C$
  • Where $P<em>A, P</em>B, P_C$ are the partial pressures of A, B, and C

Sample Problem 7.9:

• A sample of exhaled air contains four gases with the following partial pressures: $N<em>2$ (563 mm Hg), $O</em>2$ (118 mm Hg), $CO<em>2$ (30 mm Hg), and $H</em>2O$ (50 mm Hg). What is the total pressure of the sample?
  • $P<em>{total} = P</em>{N<em>2} + P</em>{O<em>2} + P</em>{CO<em>2} + P</em>{H_2O}$
  • $P_{total} = 563 + 118 + 30 + 50$
  • $P_{total} = 761 mm Hg$