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Math Help - Please help me with these chemistry problems (gas laws)?

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    Please help me with these chemistry problems (gas laws)?

    A balloon containing helium gas contains 0.32 mol He at 2432 mmHg and 25 degreesC. The volume of the helium is 2.45L. If an additional 0.14 mol of He is injected into the balloon while the temperature and pressure constant, what is the volume of the helium.

    Poisonous carbon monoxide gas is a product of the internal combustion engine. If 1.2 mol of Co at 11degreesC and 102mmHg are present in a container, what will be the volume of the CO gas?

    Assume that a sample of humid air contains only nitrogen gas, oxygen gas, and water vapor. If the atmospheric pressure is 745mmHg and the patial pressure of N2 is 566 mmHg and of oxygen is 140mmHg, what is the partial pressure of water vapor in the air.

    i am totally lost
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    Quote Originally Posted by desperate_on_sunday_night View Post
    A balloon containing helium gas contains 0.32 mol He at 2432 mmHg and 25 degreesC. The volume of the helium is 2.45L. If an additional 0.14 mol of He is injected into the balloon while the temperature and pressure constant, what is the volume of the helium.

    Poisonous carbon monoxide gas is a product of the internal combustion engine. If 1.2 mol of Co at 11degreesC and 102mmHg are present in a container, what will be the volume of the CO gas?

    Assume that a sample of humid air contains only nitrogen gas, oxygen gas, and water vapor. If the atmospheric pressure is 745mmHg and the patial pressure of N2 is 566 mmHg and of oxygen is 140mmHg, what is the partial pressure of water vapor in the air.

    i am totally lost
    From google: 2432 mmHg = 324 240 pascals

    1. Use the ideal gas law: n = \frac{PV_1}{RT_1} = \frac{324240*2.45*10^{-3}}{8.314*(25+273.15)} = 0.3205mol

    Now to find V with n_2 = n_1 + 0.14 = 0.46 mol

    V_2 = \frac{nRT}{P} = \frac{0.46*8.314*298.15}{324240} = 3.52L


    2. Again use the ideal gas law using google to convert from mmHg to Pa:

    102 torr = 13 598.8816 pascals

    V = \frac{nRT}{P} = \frac{1.2*8.314*(273.15+11)}{13599} = 0.208 m^3 = 208L


    3. Partial Pressure = mole fraction * total pressure

    mole fraction of X= moles of X/total moles

    However, in this case there is a shortcut since the total pressure must equal the sum of the partial pressures:

    p_{H_{2}O{_g}} = 745 - (566+140) = 39mmHg
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