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19 December, 03:29

A gas is contained in a thick-walled balloon. When the pressure changes from 2.95 atm to atm, the volume changes from 7.456 L to 4.782 L and the temperature changes from 379 K to 212 K

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  1. 19 December, 05:01
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    Let's assume that the gas in the balloon 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 = 2.95 atm

    V₁ = 7.456 L

    T₁ = 379 K

    P₂ = ?

    V₂ = 4.782 L

    T₂ = 212 K

    By applying the formula,

    2.95 atm x 7.456 L / 379 K = P₂ x 4.782 L / 212 K

    P₂ = (2.95 atm x 7.456 L x 212 K) / (379 K x 4.782 L)

    P₂ = 2.57 atm

    Hence, the answer is 2.57 atm.
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