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Chemistry LibreTexts

9: Gases

  • Page ID
    284427
  • Gas Properties

    Exercise \(\PageIndex{1a}\)

    A pressure of 1.00 atm has a metric equivalent of 1.01 × 105 ________.

    Answer

    Pascals

    Exercise \(\PageIndex{1b}\)

    Which of the following is the largest value of pressure?

    a. 0.716 atm        b. 18.3 in Hg        c. 972 mm Hg        d. 14.9 psi        e. 86572 Pa

    Answer

    c. 972 mm Hg

    Convert all to same units.

    a. 0.716 atm

    b. \(\large \frac{18.3 in Hg}{29.9213 (\frac{atm}{in Hg})} = 0.612\;atm\)

    c. \(\large \frac{972 mm Hg}{760 (\frac{atm}{mm Hg})}=1.28\;atm\)

    d. \(\large \frac{14.9 psi}{14.7\frac{atm}{psi}}=1.01\;atm\)

    e. \(\large \frac{86572 Pa}{1.01*10^{5}\frac{atm}{Pa}}=0.857\;atm\)

    Exercise \(\PageIndex{1c}\)

    A particular gas exerts a pressure of 356 mm Hg. What is this pressure in units of bar? (1 atm = 760 mm Hg = 101.3 kPa = 1.013 bar)

    Answer

    0.473 bar

    Exercise \(\PageIndex{1d}\)

    The local weather forecaster reports that the current barometric pressure is 15.9 inches of mercury. What is the current pressure in atmospheres?

    Answer

    0.531 atm

    Exercise \(\PageIndex{1e}\)

    It is possible to make a barometer using a liquid other than mercury. What would be the height (in meters) of a column of dichloromethane at a pressure of 0.790 atm, given that 0.790 atm is equal to a 0.758 m column of mercury and the densities of mercury and dichloromethane are 13.5 g/cm3 and 1.33 g/cm3, respectively.

    Answer

    7.69 m

    \(\large d=\frac{P_{atm}}{hg}\) can be used to find the height in a column.

    Use a ratio with the pressure in atm is equal to one. 

    \(\large \frac{P_{CH_{2}Cl_{2}}}{P_{Hg}}=\frac{d_{CH_{2}Cl_{2}}*g*h_{CH_{2}Cl_{2}}}{d_{Hg}*g*h_{Hg}}\)

    \(\large \frac{0.790}{0.790}=\frac{1.33*981*h_{CH_{2}Cl_{2}}}{13.5*981*60}\) Note: Converted to cm from m.
    \(\large h_{CH_{2}Cl_{2}}=609.02 cm\)

    Gas Laws

    Exercise \(\PageIndex{2a}\)

    Use the ideal gas law to derive an equation that relates the remaining variables for a sample of an ideal gas if the following are held constant.

    1. amount and volume
    2. pressure and amount
    3. temperature and volume
    4. temperature and amount
    5. pressure and temperature
    Answer
    1. P/T = constant
    2. V/T = constant (Charles’ law)
    3. P/n = constant
    4. PV = constant (Boyle’s law)
    5. V/n = constant (Avogadro’s law)

    Exercise \(\PageIndex{2b}\)

    A sample of gaseous Cl2 has a volume of 12.4 L at 500.0 K and 0.521 atm. How many moles are present?

    Answer

    \(n=\frac{PV}{RT}=\frac{0.521atm\left ( 12.4L \right )}{0.08206\frac{L\cdot atm}{mol\cdot K}\left ( 500K \right )}=0.157mol\)

    Exercise \(\PageIndex{2c}\)

    What is the mass of chlorine gas in a 12.4L container at 500.0K and 0.521 atm?

    Answer

    \(n=\frac{PV}{RT}=\frac{0.521atm\left ( 12.4L \right )}{0.08206\frac{L\cdot atm}{mol\cdot K}\left ( 500K \right )}=0.157mol\)

    \(0.157mol~Cl_{2}\left ( \frac{70.90g~Cl_{2}}{mol} \right )=11.1g\)

    Exercise \(\PageIndex{2d}\)

    What volume is occupied by 1.06 mol of CO2 gas at 299 K and a pressure of 0.89 atm?

    Answer

    \(V=\frac{nRT}{P}=\frac{1.06mol\left ( 0.08206\frac{L\cdot atm}{mol\cdot K} \right )299K}{0.89atm}=29L\)

    Exercise \(\PageIndex{2e}\)

    What is the temperature of 1.41 mol of methane gas in a 5.0 L container at 1.00 atm?

    Answer

    \(T=\frac{PV}{nR}=\frac{1.00atm\left ( 5.0L \right )}{1.41mol\left ( 0.08206\frac{L\cdot atm}{mol\cdot K} \right )}=43K\)

    Exercise \(\PageIndex{2f}\)

    What is the pressure exerted by a 1.75 mol sample of water at 7.0L and 20oC?

    Answer

    \(P=\frac{nRT}{V}=\frac{1.75mol\left ( 0.08206\frac{L\cdot atm}{mol\cdot K} \right )293.15K}{7.0L}=5.8atm\)

     

     

    Two State Ideal Gas Problems

    Exercise \(\PageIndex{3a}\)

    What pressure would 6.1 mole of a gas in a rigid container have at 20.0 oC have if it had a pressure of 0.45 atm at -45.0 oC?

    Answer

    \(P_{1}=P_{2}\frac{T_{1}}{T_{2}}=\left ( 0.45atm \right )\left ( \frac{293.15K}{228.15K} \right )=0.58atm\)

     

    Exercise \(\PageIndex{3b}\)

    A pressure tank containing chlorine gas at a pressure of 2.00 atm and a temperature of 40oC is set with a pressure relief valve set to open at a pressure of 10.0 atm. At what temperature will the relief valve open?

    Answer

    \(T_{2}=T_{1}\frac{P_{2}}{P_{1}}= \left ( 313.15K \right )\left ( \frac{10.0atm}{2.00atm} \right )=1570K\)

    Exercise \(\PageIndex{3c}\)

    A gas is in a sealed cylinder at 25oC with a piston which can expand or contract to change the volume. The initial volume and pressure are 75.0 L and 980.0 torr. A force is applied to the piston and the volume adjusts to a final pressure and temperature of 5.00 atm at 188oC. What is the final volume?

    Answer

    \(V_{2}=V_{1}\left ( \frac{P_{1}}{P_{2}} \right )\left ( \frac{T_{2}}{T_{1}} \right )=75.0L\left ( \frac{980torr\left ( \frac{1atm}{760torr} \right )}{5.00atm} \right )\left ( \frac{461.15K}{298K} \right )=29.9L\)

     

     

    Gas Phase Stoichiometry

    Exercise \(\PageIndex{4a}\)

    5.00 g of Mg is added to 50.0 mL of 0.800M HCl. A double displacement reaction occurs in a sealed container at 25.0oC. What is the pressure of the hydrogen gas generated if the volume above the solution is 100.0 ml, and we ignore any pressure due to the evaporated water?

    Answer

    \(P_{H_{2}}=\frac{n_{H_{2}}RT}{V}=\frac{0.0200mol~H_{2}\left ( 0.08206\frac{L\cdot atm}{mol\cdot K} \right )298K}{0.100L}=4.89atm\)

    Exercise \(\PageIndex{4b}\)

    Nitrogen dioxide is formed in a closed container at 90oC and 1.00 atm when 1.50 g NO and 2.00 mole of O2 are mixed. After the reaction is finished the pressure changes to 3.90 atm. What is the final temperature?

    Answer

    \(T_{2}=T_{1}\left ( \frac{n_{1}}{n_{2}} \right )\left ( \frac{P_{2}}{P_{1}} \right )=363.15K\left ( \frac{2.05mol}{2.025mol} \right )\left ( \frac{3.90atm}{1.00atm} \right )=1433.77K =1400K\)

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