Ideal Gas Law and Its Applications

Ideal Gas Law Introduction

• The Ideal Gas Law is represented by the equation: $PV = NRT$.

• This law describes the behavior of an ideal gas under a set of conditions.

• Key variables:

  • P = Pressure

  • V = Volume

  • N = Number of moles of gas

  • R = Universal gas constant

  • T = Temperature in Kelvin

Rearranging the Ideal Gas Law

• To isolate Pressure (P):
  $P = \frac{NRT}{V}$

• This form allows for the calculation of pressure when moles, temperature, and volume are known.

Example Problem 1: Calculating Pressure in Atmospheres

• Given:

  • Moles of gas (N) = 0.552 moles

  • Gas constant (R) = 0.08206 $L \cdot atm/(mol \cdot K)$

  • Temperature (T) = 305 K

  • Volume (V) = 8.5 L

• Calculation involves:
  $P = \frac{(0.552 \text{ moles}) \times (0.08206 \text{ L atm/(mol K)}) \times (305 \text{ K})}{8.5 \text{ L}}$

• Units show cancellation:

  • Moles cancel with moles, liters cancel with liters, Kelvin cancel with Kelvin.

  • Resulting units = atm (pressure).

• Final computation:

  • Calculated Pressure = 1.63 atm (to three significant figures).

Conversion to PSI (Pounds per Square Inch)

• To convert from atm to PSI:

  • Use the conversion factor: 1 atm = 14.7 PSI.

• Calculation:
  $1.63 \text{ atm} \times 14.7 \text{ PSI/atm} = 24 \text{ PSI}$

• Interpretation:

  • 24 PSI indicates low pressure, suitable for a bicycle tire rather than a car tire.

Example Problem 2: Solving for Volume

• Given condition:

  • Moles (N) = 0.556 moles

  • Pressure (P) = 715 mmHg (use R = 62.36 $L \cdot mmHg/(mol \cdot K)$)

  • Temperature = 58 °C (converted to Kelvin: 331.15 K)

• Rearranged Ideal Gas Law:
  $V = \frac{NRT}{P}$

• Calculation:
  $V = \frac{(0.556 \text{ moles}) \times (62.36 \text{ L mmHg/(mol K)}) \times (331.15 \text{ K})}{715 \text{ mmHg}}$

• Result: Volume = 16.1 L (to three significant figures).

Example Problem 3: Pressure of Helium Gas

• Given:

  • Mass of helium = 0.133 g

  • Volume = 648 mL (converted to 0.648 L)

  • Temperature = 32 °C (converted to Kelvin: 305.15 K)

• Convert mass to moles:

  • Molar mass of helium = 4.003 g/mol

  • Calculation: $N = \frac{0.133 \text{ g}}{4.003 \text{ g/mol}} = 0.0332 \text{ moles}$

• Ideal Gas Law setup for pressure:

  • $P = \frac{NRT}{V}$

  • Use R = 62.36 for mmHg.

• Calculation:
  $P = \frac{(0.0332 \text{ moles}) \times (62.36 \text{ L mmHg/(mol K)}) \times (305.15 \text{ K})}{0.648 \text{ L}}$

• Result: Pressure = 975 mmHg (to three significant figures).

Standard Temperature and Pressure (STP) Definition

• STP conditions define a gas's behavior:

  • Pressure = 1 atm (equivalent to 760 mmHg)

  • Temperature = 0 °C (273.15 K)

• Molar Volume:

  • 22.4 L is the volume of one mole of any ideal gas at STP.

• Avogadro's Number:

  • $6.022 \times 10^{23}$ molecules per mole.

Density of Gases

• Density of a gas: $density = \frac{mass}{volume}$.

• Density formula at STP: $d = \frac{P \times M}{R \times T}$ Where:

  • P = Pressure

  • M = Molar mass

  • R = Universal gas constant, and T = Temperature in Kelvin.

• Example Densities:

  • Helium = 0.179 g/L

  • Nitrogen = 1.25 g/L

• Explanation for Helium Balloons:

  • Helium, being less dense than air (nitrogen), allows balloons to float.

Molar Mass Calculation Example

• Density provided: 1.43 g/L, Temperature = 23 °C (296.15 K), and Pressure = 0.789 atm.

• Rearranging density equation for molar mass:
  $M = \frac{density \times R \times T}{P}$

• Plugging in values:
  $M = \frac{1.43 \text{ g/L} \times 0.08206 \text{ L atm/(mol K)} \times 296.15 \text{ K}}{0.789 \text{ atm}}$

• Result: Molar mass = 44 g/mol, suggesting CO2 as the gas in question.

Conclusion

• Understanding the combinations and manipulations of ideal gases using PV = NRT allows for various practical applications including pressure, volume, temperature, and density calculations.

• Ideal Gas Law demonstrates the predictable behavior of gases when measured under standard conditions and facilitates conversions between different pressure units (e.g., atm to PSI, mmHg).