Ideal Gas Law Notes

Chemical Reactions and Phases

• Chemical reactions occur between different substances with different phases.
• Examples:
  • $H<em>2O(g) + 5O</em>2(g) \rightarrow H<em>r SO</em>y(my)$
  • $2Ca(s) + O_2(g) \rightarrow 2CaO(s)$

States of Matter

• Solid: Definitive shape and volume.
• Liquid: Definite volume, adopts the shape of its container. Incompressible.
• Gas: No definite shape or volume; fills any container and is compressible.
  • Volume depends on other properties.
• Particles in a gas are far apart with little interaction.

Ideal Gas Definition

• Particles have no attraction to each other.
• They collide like billiard balls.
• Volume of particles is much smaller than the volume of the container.
• Many gases can be approximated as ideal.

Gas Properties and Units

• Volume: 3D size gas occupies, same as the size of the container.
  • Particles are always moving and fill the container space.
  • Units: L, mL, $cm^3$, $m^3$, etc.
• Temperature: Measure of kinetic energy of gas particles.
  • Depends on the speed (and mass) of particles.
  • Slow = cold, Fast = hot.
  • Units: °C or Kelvin (K).
  • $K = °C + 273.15$
  • Absolute temperature scale: no negative temperatures in K.
  • Absolute zero: all motion ceases.
  • A change of 1°C = change of 1K, but 1°C ≠ 1K.
  • Example: 20°C → 30°C is the same temperature change as 293K → 303K.

Volume Conversion Examples

• Convert 785 $cm^3$ to L:
  • $785 cm^3 * (\frac{1 L}{1000 cm^3}) = 0.785 L$
• Convert 2.2 $m^3$ to L:
  • $2.2 m^3 * (\frac{100 cm}{1 m})^3 * (\frac{1 L}{1000 cm^3}) = 2.2 * 10^3 L = 2200 L$

Temperature Conversion Example

• Convert -65°C to Kelvin:
  • $-65°C + 273.15 = 208.15 K$

Pressure

• Measure of the "push" particles exert on the walls of the container due to collisions.
• $P = \frac{force}{area}$
• Air pressure is the force of air pressing down on a certain area.
• Measured using a barometer, often with liquid mercury.
• Units: mm Hg (torr), atmosphere (atm), kilopascals (kPa), bars, PSI.

Pressure Conversions

• 1 mm Hg = 1 torr
• 1 atm = 760 mm Hg
• 1 atm = 101.3 kPa
• 1 atm = 14.70 PSI
• 1 atm = 1.013 bar

Pressure Conversion Examples

• Air pressure at the summit of Mt. Everest is 253 mm Hg. Convert to atm, psi, kPa, and bar.
  • $253 mmHg * (\frac{1 atm}{760 mmHg}) = 0.333 atm$
  • $253 mmHg * (\frac{14.7 PSI}{760 mmHg}) = 4.90 psi$
  • $253 mmHg * (\frac{101.3 kPa}{760 mmHg}) = 33.7 kPa$
  • $253 mmHg * (\frac{1.013 bar}{760 mmHg}) = 0.337 bar$
• Air pressure on the surface of Venus is 93 bar. Convert to atm, mm Hg and kPa.
  • $93 bar * (\frac{1 atm}{1.013 bar}) = 91.8 atm$
  • $93 bar * (\frac{760 mmHg}{1.013 bar}) = 7.0 * 10^4 mmHg$
  • $93 bar * (\frac{101.3 kPa}{1.013 bar}) = 9335 kPa$

Standard Temperature and Pressure (STP)

• Defined as:
  • 0°C (273.15 K)
  • 1 atm
• Often problems state "STP" rather than give values.

Ideal Gas Law

• $PV = nRT$
  • P = pressure
  • V = volume
  • n = number of moles of gas
  • R = gas constant = 0.0821 L atm / (mol K)
  • T = temperature
• Units of P, V, must match units of R.

Ideal Gas Law Example

• A portable oxygen tank has a volume of 2.40 L and a pressure of 243 psi at a temperature of 22°C. How many moles of oxygen are present? What is the mass of oxygen?
  • Given: V = 2.40 L, P = 243 psi, T = 22°C
  • Convert P to atm: $243 psi * (\frac{1 atm}{14.7 psi}) = 16.5 atm$
  • Convert T to K: $22°C + 273.15 = 295 K$
  • $n = \frac{PV}{RT} = \frac{(16.5 atm)(2.40 L)}{(0.0821 \frac{L atm}{mol K})(295 K)} = 1.64 mol$
  • Mass of oxygen: $1.64 mol * (\frac{32.00 g}{1 mol}) = 52.5 g$

Ideal Gas Law Example 2

• At what temperature does 1.20 moles of hydrogen gas occupy a volume of 28.1 L at a pressure of 121 kPa?
  • Convert P to atm: $121 kPa * (\frac{1 atm}{101.3 kPa}) = 1.19 atm$
  • $T = \frac{PV}{nR} = \frac{(1.19 atm)(28.1 L)}{(1.20 mol)(0.0821 \frac{L atm}{mol K})} = 339 K$
  • Convert to Celsius: $339 K - 273.15 = 66°C$

Ideal Gas Law Example 3

• A room with a volume of 50 $m^3$ has a pressure of 750 torr at a temperature of 21°C. How many moles of gas occupy the room?
  • Convert V to L: $50 m^3 * (\frac{1000 L}{1 m^3}) = 50000 L$
  • Convert P to atm: $750 torr * (\frac{1 atm}{760 torr}) = 0.987 atm$
  • Convert T to K: $21°C + 273.15 = 294 K$
  • $n = \frac{PV}{RT} = \frac{(0.987 atm)(50000 L)}{(0.0821 \frac{L atm}{mol K})(294 K)} = 2025 mol \approx 2000 mol$

Relationships Among Quantities (Constant n and T)

• Suppose a container with a moveable top (piston) keeps n and T constant, change the volume. How does P change?
• $P<em>1V</em>1 = nRT = P<em>2V</em>2 = nRT$
• Boyle's Law: $P<em>1V</em>1 = P<em>2V</em>2$

Relationships Among Quantities (Constant P)

• If pressure is kept constant and volume is changed, what happens to temperature?
• Charles's Law: $\frac{V<em>1}{T</em>1} = \frac{nR}{P} \Leftrightarrow \frac{V<em>2}{T</em>2} = \frac{nR}{P}$
• $\frac{V<em>1}{T</em>1} = \frac{V<em>2}{T</em>2}$

Combined Gas Law

• When all three (P, V, and T) change and n is constant:
• $\frac{P<em>1V</em>1}{T<em>1} = nR \Leftrightarrow \frac{P</em>2V<em>2}{T</em>2} = nR$
• $\frac{P<em>1V</em>1}{T<em>1} = \frac{P</em>2V<em>2}{T</em>2}$

Using Ideal Gas Law

• Don't have to memorize these gas laws. Just pay attention to what stays constant.

Example: Volume Change with Temperature

• A balloon has a volume of 3.2 L at 25°C. The gas in the balloon is heated to 100°C. What is the new volume?
• Moles and pressure are not stated, so assume they are constant.
• $\frac{V}{T} = nR = constant$
• $T_1 = 25°C + 273.15 = 298 K$
• $T_2 = 100°C + 273.15 = 373 K$
• $V<em>2 = \frac{V</em>1T<em>2}{T</em>1} = \frac{(3.2 L)(373 K)}{(298 K)} = 4.0 L$

Example: Volume Change with Pressure

• A compressor stores 2.8 L of air at a pressure of 150 psi. If this is allowed to expand until the pressure is 15 psi, what volume will the air occupy?
• Moles and T not stated, so assume constant
• $P<em>1V</em>1 = P<em>2V</em>2$
• $V<em>2 = \frac{P</em>1V<em>1}{P</em>2}$
• $150 psi = 150 psi * (\frac{1 atm}{14.7 psi}) = 10.2 atm$
• $15 psi = 15 psi * (\frac{1 atm}{14.7 psi}) = 1.02 atm$
• $V_2 = \frac{(10.2 atm)(2.8 L)}{(1.02 atm)} = 28 L$

Example: Pressure Change with Temperature

• On a cold morning when the temperature is -10°C, a truck tire is inflated to 45.0 psi. It is then driven south until the temperature is a balmy 25°C. What is the pressure in the tire? Assume $n$ and $V$ are constant.
• $T_1 = -10°C + 273.15 = 263 K$
• $T_2 = 25°C + 273.15 = 298 K$
• $P_1 = 45.0 psi = 45.0 psi * (\frac{1 atm}{14.7 psi}) = 3.06 atm$
• $\frac{P<em>1}{T</em>1} = \frac{P<em>2}{T</em>2}$
• $P<em>2 = \frac{P</em>1T<em>2}{T</em>1} = \frac{(3.06 atm)(298 K)}{(263 K)} = 3.47 atm$
• $P_2 = 3.47 atm * (\frac{14.7 psi}{1 atm}) = 51 psi$

Example: Combined Gas Law

• A gas at a temperature of 7.0°C and pressure 200 kPa occupies a volume of 25.8 L. If the gas is compressed to three-fourths its original volume, the pressure increases to 350 kPa. What is the new temperature?
• $T_1 = 7.0 + 273 = 280 K$
• $V_1 = 25.8 L$
• $V_2 = \frac{3}{4} * 25.8 L = 19.35 L$
• $P_2 = 350 kPa = 350 kPa * (\frac{1 atm}{101.3 kPa}) = 3.46 atm$
• $P_1 = 200 kPa = 200 kPa * (\frac{1 atm}{101.3 kPa}) = 1.97 atm$
• $\frac{P<em>1V</em>1}{T<em>1} = \frac{P</em>2V<em>2}{T</em>2}$
• $T<em>2 = \frac{T</em>1P<em>2V</em>2}{P<em>1V</em>1} = \frac{(280 K)(3.46 atm)(19.35 L)}{(1.97 atm)(25.8 L)} = 368.8 K \approx 370 K$
• $T_2 = 370 K - 273.15 = 95°C$

Stoichiometry and Ideal Gas Law Example

• How many liters of oxygen gas at 21°C and 1.13 atm can be produced by heating 0.950 g $KClO<em>3$ according to: $2KClO</em>3(s) \rightarrow 2KCl(s) + 3O_2(g)$
• Use stoichiometry to find the number of moles of $O_2$ produced:
  • $moles O<em>2 = 0.950 g KClO</em>3 * (\frac{1 mol KClO<em>3}{122.6 g KClO</em>3}) * (\frac{3 mol O<em>2}{2 mol KClO</em>3}) = 0.01163 mol O_2$
• Then, use the ideal gas law to find the volume of $O_2$:
  • $V = \frac{nRT}{P} = \frac{(0.01163 mol O<em>2)(0.0821 \frac{L atm}{mol K})(21 + 273) K}{1.13 atm} = 0.248 L O</em>2$