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Calculate the volume occupied by 4.0 g of oxygen at STP, and the pressure exerted by 18.5 g of nitrogen in a 5 L container at 27°C.

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4 parts
9

2.8 L2.8 \text{ L}

10

3.30×105 Pa3.30 \times 10^5 \text{ Pa}

11

6.21×10−21 J6.21 \times 10^{-21} \text{ J}

12

1397.12 m/s1397.12 \text{ m/s}

Step-by-step solution

10 steps

Part 9· 3 steps

  1. Identify the given values and conditions.

    Mass of oxygen m=4.0 gm = 4.0 \text{ g}.
    Molar mass of oxygen M=32 g/molM = 32 \text{ g/mol}.
    The conditions are S.T.P (Standard Temperature and Pressure).
    At S.T.P., T=0∘C=273.15 KT = 0^\circ \text{C} = 273.15 \text{ K} and P=1 atm=101325 PaP = 1 \text{ atm} = 101325 \text{ Pa}.
    Also, at S.T.P., 1 mole of any ideal gas occupies 22.4 L22.4 \text{ L}.

  2. Calculate the number of moles (nn) of oxygen.

    n=mM=4.0 g32 g/moln = \frac{m}{M} = \frac{4.0 \text{ g}}{32 \text{ g/mol}} n=0.125 moln = 0.125 \text{ mol}

  3. Calculate the volume occupied by the oxygen at S.T.P.

    Since 1 mole occupies 22.4 L22.4 \text{ L} at S.T.P.:

    V=n×22.4 L/molV = n \times 22.4 \text{ L/mol} V=0.125 mol×22.4 L/molV = 0.125 \text{ mol} \times 22.4 \text{ L/mol} V=2.8 LV = 2.8 \text{ L}

    The volume occupied by 4.0 g4.0 \text{ g} of oxygen at S.T.P is 2.8 L\boxed{2.8 \text{ L}}.

Part 10

3.30×105 Pa3.30 \times 10^5 \text{ Pa}

  1. Identify the given values and convert them to SI units.

    Mass of nitrogen m=18.5 gm = 18.5 \text{ g}.
    Volume V=5 L=5×10−3 m3V = 5 \text{ L} = 5 \times 10^{-3} \text{ m}^3.
    Temperature T=27∘C=27+273=300 KT = 27^\circ \text{C} = 27 + 273 = 300 \text{ K}.
    Molar mass of nitrogen M=28 g/mol=0.028 kg/molM = 28 \text{ g/mol} = 0.028 \text{ kg/mol}.
    Universal Gas Constant R=8.314 J/(mol⋅K)R = 8.314 \text{ J/(mol}\cdot\text{K)}.

  2. Calculate the number of moles (nn) of nitrogen.

    n=mM=18.5 g28 g/moln = \frac{m}{M} = \frac{18.5 \text{ g}}{28 \text{ g/mol}} n≈0.6607 moln \approx 0.6607 \text{ mol}

  3. Apply the Ideal Gas Law to find the pressure PP.

    The Ideal Gas Law is PV=nRTPV = nRT. Rearrange to solve for PP:

    P=nRTVP = \frac{nRT}{V} P=(0.6607 mol)×(8.314 J/(mol⋅K))×(300 K)5×10−3 m3P = \frac{(0.6607 \text{ mol}) \times (8.314 \text{ J/(mol}\cdot\text{K)}) \times (300 \text{ K})}{5 \times 10^{-3} \text{ m}^3} P=1647.95×10−3 PaP = \frac{1647.9}{5 \times 10^{-3}} \text{ Pa} P=329580 PaP = 329580 \text{ Pa} P≈3.30×105 PaP \approx 3.30 \times 10^5 \text{ Pa}

    The pressure of the nitrogen gas is 3.30×105 Pa\boxed{3.30 \times 10^5 \text{ Pa}}.

Part 11

6.21×10−21 J6.21 \times 10^{-21} \text{ J}

  1. Identify the given values and convert temperature to Kelvin.

    Temperature T=27∘C=27+273=300 KT = 27^\circ \text{C} = 27 + 273 = 300 \text{ K}.
    Boltzmann constant k=1.38×10−23 J/Kk = 1.38 \times 10^{-23} \text{ J/K}.

  2. Calculate the average translational kinetic energy (KEavgKE_{avg}) of a nitrogen molecule.

    The formula for the average translational kinetic energy per molecule is:

    KEavg=32kTKE_{avg} = \frac{3}{2} kT KEavg=32×(1.38×10−23 J/K)×(300 K)KE_{avg} = \frac{3}{2} \times (1.38 \times 10^{-23} \text{ J/K}) \times (300 \text{ K}) KEavg=1.5×1.38×10−23×300 JKE_{avg} = 1.5 \times 1.38 \times 10^{-23} \times 300 \text{ J} KEavg=621×10−23 JKE_{avg} = 621 \times 10^{-23} \text{ J} KEavg=6.21×10−21 JKE_{avg} = 6.21 \times 10^{-21} \text{ J}

    The average translational kinetic energy of a nitrogen molecule at 27∘C27^\circ \text{C} is 6.21×10−21 J\boxed{6.21 \times 10^{-21} \text{ J}}.

Part 12

1397.12 m/s1397.12 \text{ m/s}

  1. Identify the given values and convert them to SI units.

    Temperature T=40∘C=40+273=313 KT = 40^\circ \text{C} = 40 + 273 = 313 \text{ K}.
    Molar mass of helium M=4.0 Kg/KmoleM = 4.0 \text{ Kg/Kmole}. This is equivalent to 4.0 g/mol4.0 \text{ g/mol} or 0.004 kg/mol0.004 \text{ kg/mol}.
    Universal Gas Constant R=8.314 J/(mol⋅K)R = 8.314 \text{ J/(mol}\cdot\text{K)}.

  2. Calculate the root mean square speed (vrmsv_{rms}).

    The formula for the root mean square speed is:

    vrms=3RTMv_{rms} = \sqrt{\frac{3RT}{M}} vrms=3×(8.314 J/(mol⋅K))×(313 K)0.004 kg/molv_{rms} = \sqrt{\frac{3 \times (8.314 \text{ J/(mol}\cdot\text{K)}) \times (313 \text{ K})}{0.004 \text{ kg/mol}}} vrms=7807.740.004 m/sv_{rms} = \sqrt{\frac{7807.74}{0.004}} \text{ m/s} vrms=1951935 m/sv_{rms} = \sqrt{1951935} \text{ m/s} vrms≈1397.12 m/sv_{rms} \approx 1397.12 \text{ m/s}

    The root mean square speed of a helium atom at 40∘C40^\circ \text{C} is 1397.12 m/s\boxed{1397.12 \text{ m/s}}.

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