Day 3: Combined and Ideal Gas Law

Combined Gas Law

  • The Combined Gas Law derives from the relationship between pressure (P), volume (V), and temperature (T) of a gas, keeping the amount of gas constant.

  • Defined by the equation:
    P1×V1T1=P2×V2T2\frac{P1 \times V1}{T1} = \frac{P2 \times V2}{T2}

  • Variables Explained:

    • P = Pressure (in mmHg)

    • V = Volume (in mL)

    • T = Temperature (in Kelvin)

    • Subscripts:

    • 1 indicates initial conditions

    • 2 indicates final conditions

Example Problem Using Combined Gas Law

  • Given:

    • Initial volume (V1) = 1.58 mL

    • Initial pressure (P1) = 735 mmHg

    • Initial temperature (T1) = 34 °C (Convert to Kelvin: T1 = 307 K)

    • Final volume (V2) = 1.08 mL

    • Final temperature (T2) = 85 °C (Convert to Kelvin: T2 = 358 K)

    • Final pressure (P2) is unknown.

  • With known values, the equation setup becomes:
    735 mmHg×1.58 mL307 K=P2×1.08 mL358 K\frac{735 \text{ mmHg} \times 1.58 \text{ mL}}{307 \text{ K}} = \frac{P_2 \times 1.08 \text{ mL}}{358 \text{ K}}

  • Calculation Steps:

    • Cross multiply to isolate P2:
      P2×1.08 mL=735×1.58×358307P_2 \times 1.08 \text{ mL} = \frac{735 \times 1.58 \times 358}{307}

    • Estimate the resulting final pressure, which will be greater than the initial pressure due to compression and temperature rise.

    • After calculation, it yields:
      P2=1253.9 mmHg(rounded to 1250 mmHg)P_2=1253.9\text{ mmHg}\quad(\text{rounded to }1250\text{ mmHg})

Ideal Gas Law

  • The Ideal Gas Law combines the principles of earlier gas laws into one equation:
    PV=nRTPV = nRT

  • Variables Explained:

    • P = Pressure (in atm)

    • V = Volume (in L)

    • n = Number of moles of gas

    • R = Ideal gas constant = 0.0821 L atm/(mol K)

    • T = Temperature (in Kelvin)

  • Unit Consideration:

    • Volume must be in liters

    • Pressure must be in atmospheres

    • Temperature in Kelvin

Example Problem Using Ideal Gas Law

  • Calculate the Volume Occupied by Nitrogen Gas:

    • Given:

    • Moles (n) = 0.854 moles

    • Pressure (P) = 1.37 atm

    • Temperature (T) = 315 K

  • Setup the Equation:
    V=nRTPV = \frac{nRT}{P}

  • Calculating Volume:

    • Substitute known values into the equation:
      V=0.854 mol×0.0821 L atm/(extmolK)×315 K1.37 atmV = \frac{0.854 \text{ mol} \times 0.0821 \text{ L atm} / ( ext{mol K}) \times 315 \text{ K}}{1.37 \text{ atm}}

    • Final calculation yields:
      V=16.0 LV = 16.0 \text{ L}

Conversion and Final Calculation for Helium Sample

  • Given:

    • Mass = 0.133 g of helium

    • Volume = 648 mL (Convert to liters: V = 0.648 L)

    • Temperature = 32 °C (Convert to Kelvin: T = 305 K)

  • Convert Mass to Moles:

    • Molar mass of Helium = 4 g/mol

    • Moles (n) = 0.133 g ÷ 4 g/mol = 0.03325 moles

  • Setup the Ideal Gas Law for Pressure (P):
    P=nRTVP = \frac{nRT}{V}

  • Substitute Values into the Pressure Equation:
    P=0.03325 mol×0.0821 L atm/(extmolK)×305 K0.648 LP = \frac{0.03325 \text{ mol} \times 0.0821 \text{ L atm} / ( ext{mol K}) \times 305 \text{ K}}{0.648 \text{ L}}

    • This results in pressure in atmospheres:
      P=1.28 atmP = 1.28 \text{ atm}

  • Convert Pressure to mmHg:

    • 1 atm = 760 mmHg

    • PmmHg=1.28 atm×760 mmHg/atm=976.5 mmHgP_{mmHg} = 1.28 \text{ atm} \times 760 \text{ mmHg/atm} = 976.5 \text{ mmHg}

    • Final pressure rounded to significant figures yields:
      P=977 mmHgP = 977 \text{ mmHg}