Hey there! Sign in to join this conversationNew here? Join for free
    • Thread Starter
    Offline

    2
    ReputationRep:
    A nebula−a region of the galaxy where new stars are forming−contains a very tenuous gas with 100 atoms/cm3. This gas is heated to 7500 K by ultraviolet radiation from nearby stars.
    What is the pressure, in atm?

    I'm having troubles converting between units. So far my work is:
    pV=nRT so p=nRT/V

    Use avogadro's number to find n=1.66x10^-22

    p=(1.66x10^-22mol)(8.314J/mol K)(7500K)/(1x0^-4m^3)
    p=1.035x10^-13Pa

    This is where I start to get confused.
    1atm=101325Pa
    9.87x10^-6atm=1Pa
    1.035x10^-13Pa=1.02x10^-18atm.
    Is this correct or is there something wrong in the conversions between units?
    Offline

    20
    ReputationRep:
    (Original post by PatchworkTeapot)
    A nebula−a region of the galaxy where new stars are forming−contains a very tenuous gas with 100 atoms/cm3. This gas is heated to 7500 K by ultraviolet radiation from nearby stars.
    What is the pressure, in atm?

    I'm having troubles converting between units. So far my work is:
    pV=nRT so p=nRT/V

    Use avogadro's number to find n=1.66x10^-22

    p=(1.66x10^-22mol)(8.314J/mol K)(7500K)/(1x0^-4m^3)
    p=1.035x10^-13Pa

    This is where I start to get confused.
    1atm=101325Pa
    9.87x10^-6atm=1Pa
    1.035x10^-13Pa=1.02x10^-18atm.
    Is this correct or is there something wrong in the conversions between units?
    number of 1 cm cubes in 1 m cube = (number of cm in 1 m)3 and != 1.00x10^4
    • Thread Starter
    Offline

    2
    ReputationRep:
    (Original post by Joinedup)
    number of 1 cm cubes in 1 m cube = (number of cm in 1 m)3 and != 1.00x10^4
    So does that mean that the overall pressure is

    p=(1.66x10^-22mol)(8.314J/mol K)(7500K)/(1x10^4m^3)
    p=1.035x10^-21Pa
    p=1.02x10^-26atm?
    Offline

    20
    ReputationRep:
    (Original post by PatchworkTeapot)
    So does that mean that the overall pressure is

    p=(1.66x10^-22mol)(8.314J/mol K)(7500K)/(1x10^4m^3)
    p=1.035x10^-21Pa
    p=1.02x10^-26atm?
    moles of gas in 1cm^3 = 1.66e-22
    R=8.314
    T=7500
    V=1/(100^3) = 1.00e-6

    p=nRT/V
    p=(1.66e-22) (8.314) (7500) (1/1.00e-6)
    =1.04e-11 (Pascals)
    =1.03e-16 (atmos)
 
 
 
  • See more of what you like on The Student Room

    You can personalise what you see on TSR. Tell us a little about yourself to get started.

  • Poll
    What's your favourite Christmas sweets?
  • See more of what you like on The Student Room

    You can personalise what you see on TSR. Tell us a little about yourself to get started.

  • The Student Room, Get Revising and Marked by Teachers are trading names of The Student Room Group Ltd.

    Register Number: 04666380 (England and Wales), VAT No. 806 8067 22 Registered Office: International House, Queens Road, Brighton, BN1 3XE

    Quick reply
    Reputation gems: You get these gems as you gain rep from other members for making good contributions and giving helpful advice.