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3 January, 23:52

A marshmallow has an initial volume of 0.084 L at standard pressure (1.0 atm). If the marshmallow is placed in a vacuum chamber and the final volume is 0.785L, what is the pressure inside the chamber?

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  1. 4 January, 00:44
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    The new pressure on the gas P2 = 0.10atm

    Explanation:

    Data;

    V1 = 0.084L

    P1 = 1.0atm

    V2 = 0.785L

    P2 = ?

    This question is a practical problem where Boyle's law is applied.

    According to Boyle's law, the pressure of a fixed mass of gas is inversely proportional to its volume provided that the temperature on the gas remains constant

    Mathematically

    P = k / v, k = PV

    P1*V1 = P2*V2 = P3*V3 ... Pn*Vn

    P1 * V1 = P2 * V2

    Solving for P2,

    P2 = (P1 * V1) / V2

    P2 = (0.084 * 1) / (0.785)

    P2 = 0.10atm

    The new pressure of the gas is 0.10atm
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