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26 December, 05:42

If a gas occupies 30 l at 1.3 atm and 300 k what volume will it occupy at stp

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Answers (2)
  1. 26 December, 05:51
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    Let's assume that the given gas is an ideal gas.

    We can use combined gas law,

    PV/T = k (constant)

    Where, P is the pressure of the gas, V is volume of the gas and T is the temperature of the gas in Kelvin.

    For two situations, we can use that as,

    P₁V₁/T₁ = P₂V₂/T₂

    P₁ = 1.3 atm

    V₁ = 30 L

    T₁ = 300 K

    P₂ = Standard pressure = 1 atm

    V₂ = ?

    T₂ = Standard temperature = 273 K

    By applying the formula,

    1.3 atm x 30 L / 300 K = 1 atm x V₂ / 273 K

    V₂ = (1.3 atm x 30 L x 273 K) / (300 K x 1 atm)

    V₂ = 35.49 L

    Hence, the volume of the gas at STP is 35.49 L.
  2. 26 December, 07:40
    0
    Answer : Using t he ideal gas equation, which is;

    PV=nRT.

    Where, P = 1.3 atm, T = 300K.

    We can consider, n and R are constant.

    now, on comparing the gas at STP conditions (At standard pressure and temperature),

    P′ = 1 atm, and T′ = 273K;

    V′ = ?

    So, the ideal gas equation is modified as P′V′=nRT′.

    Dividing the equations, P′/P X V′/V=T′/T

    Therefore, (1 atm / 1.3 atm) X (V' / 30 L) = (273 K / 300 K)

    V' = ~ 20.9 L

    So, the volume of gas at STP will be ~ 20.9 L
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