Standard Temperature and Pressure?

Im studying for a chem test and i dont understand STP, can you explain how to do this problem:

At 27 C and 800 mmHg pressure, a quantity of hydrogen has a volume of 22 mL. what would the volume be under standard conditions?

4 Answers

  • Anonymous
    1 decade ago
    Favorite Answer

    at STP, Pressure is 1atm, and Temperature is 273K. ALWAYS.

    use (P)(V) / (t)= (P1)(V1) / (T1).

    convert 27C to K. 27+273 = 300K.

    convert 800mmHG to atm. 800/760=1.053atm.

    convert ml to L. 22mL= .022L.

    insert into the equation: (1.053atm)(.022)/300K = (1atm)(x L)/(273K)

    solve for x, which is your volume.

    (1.053atm)(.022)/300K = (1atm)(x L)/(273K)

    ==0.023166/300 = x/273



    so your new volume is .02108 L, or 21 mL.

  • Norrie
    Lv 7
    1 decade ago

    Standard conditions for gases are taken as 273 Kelvin (0°C) and 1.0 atmosphere of Pressure taken at sea level.

    1.0 atm. is also equal to 760mmHg (or 760 torr); and, 101.325 kPa. (In Imperial, it's 14.7 psia).

    For your question, the Combined Gas Law can be used...

    (P1 x V1 x T2 = P2 x V2 x T1).

    Pressure conversion to atmospheres and mL conversion to Litres, is not necessary.

    P1 = 800mmHg; V1 = 22mL; T2 = 273K.

    P2 = 760mmHg; V2 = ? and T1 = (27°C + 273) = 300K

    Plug these values into the calculation.

    800mmHg x 22mL x 273K = 760mmHg x V2 x 300K.

    V2 (new volume) = (800 x 22 x 273) / (760 x 300).

    = 4,804,800 / 228,000 = 21.07mL final volume at STP.

    Source(s): Senior Technical Instructor.
  • Anonymous
    4 years ago

    -273 C is 0 ok, which isn't an pretty popular temperature. no rigidity is likewise no longer an pretty popular rigidity. by potential of technique of removing, it is going to be D. i in my opinion did no longer understand that 0 C became the popular temp; i presumed it became 25 C. You learn something every day!

  • 1 decade ago

    STP is defined as 0C (= 273C) and 1atm (= 760mmHg)

    Using the combined gas law (P1V1/T1 = P2V2/T2)

    800mmHg * .022L / 300K = 760mmHg * xL / 273K

    x = .021L = 21mL

    Source(s): I am an AP Chemistry student at Thomas Jefferson High School for Science and Technology.
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