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12 May, 03:21

If 20 ml of gas is subjected to a temperature change from 10°C to 100°C and a pressure change from 1 atm to 10 atm what's the best expression representing the new volume

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  1. 12 May, 07:20
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    V1 = 20 ml=0.020 L

    T1 = 10⁰+273⁰ = 283 K

    T2 = 100⁰ + 273⁰ = 373 K

    P1 = 1 atm

    P2 = 10 atm

    V2-?

    P1V1/T1 = P2V2/T2

    (1 atm * 0.020 L) / 283K = 10 atm*V2/373K

    V2 = (1 atm * 0.020 L*373K) / (283K * 10 atm) =

    (1 * 0.020 * 373) / (283 * 10) = 0.00264 L ≈ 2.6 ml
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