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Gas is confined in a tank at a pressure of 10.0 atm and a temperature of 15o
C. If
half of the gas is withdrawn and the temperature is raised to 65o
C, what is the new
pressure of the gas?

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Angie

  1. Pi = Initial Pressure (Atmospheres)
    Pf = Final pressure (Atmospheres)
    V = Volume
    n = Moles
    R = That constant number
    Ti = Initial Temperature (Kelvins)
    Tf = Final Temperature (Kelvins)

    Use the Ideal Gas law.
    PiV=nRTi & PfV=nRTf
    ----In the problem, half the gas is removes, so there will be half the moles of gas. If we solve both equations for N, we get:
    n=PiV/RTi & n=2PfV/RTf
    -----Notice the 2nd equation is multiplied by two in order to keep the same n value as the first equation. This is so we can combine the equations and solve for Pf:
    PiV/RTi=2PfV/RTf
    ----We can cancel out R and V as they keep the same place and value in each equation.
    Pi/Ti=2Pf/Tf
    ----Now let's isolate the value we want to find. Final pressure
    Pf=Pi*Tf/Ti*2
    ----
    Plug in the values and solve it. Kelvin must be used.
    Pf=(10atm)(338.15K)/(288.15K)(2)
    Pf=(3381.5)/(576.3)
    Pf=5.867603679...... Atmospheres

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    nate

  2. you need to multiply by 0.5 not by 2

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    Jack

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