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18 April, 06:16

A sample of sulfur dioxide (SO2) is initially at a temperature of 133 C a volume of 20 L, and a pressure of 850 mm Hg. If the volume changes to 25 L and the temperature increases to 181 C, what is the new pressure?

Show your work. Don't forget to convert C to kelvin

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  1. 18 April, 07:12
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    760.39 mmHg (approx.)

    Explanation:

    Okay so given the first set of information, we have:

    T = 406 K (133+273)

    V = 20 L

    P = 1.1184 atm (converted for continuity reasons from mmHg)

    R = 0.08206 L atm/mol K

    Using PV = nRT (ideal gas law), we solve for n (moles of SO2).

    n = PV/RT = (1.1184 atm * 20 L) / (0.08206 L atm/mol K * 406 K) = 0.6714 mol

    Now that you have all the information, you can use the new volume and temperature to solve for pressure.

    T = 454 K (181+273)

    V = 25 L

    n = 0.6714 mol

    R = 0.08206 L atm/mol K

    Using the ideal gas law again, we solve for P (pressure in atm).

    P = nRT/V = (0.6714 mol * 0.08206 L atm/mol K * 454 K) / (25 L) = 1.0005 atm

    Then we convert atm back to mmHg to get 760.39 mmHg.
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