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Chemistry and gas law problems

  1. Apr 16, 2012 #1
    Hello. If somebody has time can you check my answers?

    Q1: A gas occupies a 2.0L container at 25°C and 2.0atm. If the volume of the container is halved and the temperature remains constant, what is the new pressure?

    P1V1 = P2V2

    P2 = 2.00 atm x (2.00L / 1.00L) = 4.00 atm

    ---

    Q2: The volume of 50mL of an ideal gas at STP increases to 100mL. If the pressure remains constant, what is the new temp?

    V1/T1 = V2/T2

    T2 = (100mL / 50 mL) x 273K
    T2 = 546K

    ---

    Q3: How many moles of Oxygen gas, O2, are present in a 3.5L container held at a pressure of 0.5atm and a temp of 77°C?

    n = PV/RT

    n = (0.5atm x 3.5L) / (0.082 L•atm/mol•K x 350K)
    n = 50.225 mole O2

    Thanks in advanced if you take the time to check these for me.
     
  2. jcsd
  3. Apr 20, 2012 #2
    Your method is correct, I didn't check your math. You have a calculator for that!

    Good luck!
     
  4. Apr 21, 2012 #3

    Borek

    User Avatar

    Staff: Mentor

    This MUST be wrong. At STP 1 mole of gas occupies 22.4 L. 0.5atm and 77°C is not STP, but it is also not that far - so 3.5 L should be just a fraction of a mole, not 50 moles.
     
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