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7 July, 18:38

If 0.214 mol of argon gas occupies a volume of 343.4 mL at a particular temperature and pressure, what volume would 0.375 mol of argon gas occupy under the same conditions?

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  1. 7 July, 18:53
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    601.8mL

    Explanation:

    First, we'll begin by obtaining an equation connecting volume and number of mole together. This is illustrated below:

    From the ideal gas equation:

    PV = nRT

    Divide both side by P

    V = nRT/P

    Now divide both side by n

    V/n = RT/P

    Under the same conditions simply means temperature (T), pressure (P) and the gas constant (R) are all constant i. e

    RT/P = > constant

    Therefore, the above equation can be written as:

    V1/n1 = V2/n2

    Now we can solve for the final volume otherwise known as the new volume as follow:

    Data obtained from the question include:

    n1 (initial mole) = 0.214 mol

    V1 (initial volume) = 343.4 mL

    n2 (final mole) = 0.375 mol

    V2 (final volume) = ?

    Applying the the equation V1/n1 = V2/n2, the final volume is obtain as illustrated below:

    V1/n1 = V2/n2

    343.4/0.214 = V2/0.375

    Cross multiply to express in linear form

    0.214 x V2 = 343.4 x 0.375

    Divide both side by 0.214

    V2 = (343.4 x 0.375) / 0.214

    V2 = 601.8mL

    Therefore, 0.375 mol of argon gas will occupy 601.8mL under the same condition.
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