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24 February, 21:33

760 torr = 101.3 kPa = 14.7 psi = 1 atm

A machine has 30.0 L of helium kept at - 268ºC and a pressure of 207 torr. What would the new volume be if the helium is allowed to warm to 20.0ºC and 101.3 kPa?

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  1. 24 February, 22:05
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    (P1) (V1) / (T1) = (P2) (V2) / (T2)

    Set up as: (P1) (V1) (T2) / (T1) (P2) = (V2)

    P1=207torr

    V1=30.0L

    T1=-268°C

    P2=101.3kPa (Same as 760torr) (use torr to make it easier though)

    T2=20.0°C

    Make sure all °C is in Kelvin

    °C + 273 = Kelvin

    Numbers are now:

    P1=207torr

    V1=30.0L

    T1=5K

    P2=760torr

    T2=293K

    Set them up as shown above

    Should look like this:

    (207torr) (30.0L) (293K) / (5K) (760torr) = V2

    As long as you do them in that order you should get: 478.82L

    You can also round to get 479L

    Either one is correct.
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