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23 October, 01:20

4.2 mol of monatomic gas A interacts with 3.2 mol of monatomic gas B. Gas A initially has 9500 J of thermal energy, but in the process of coming to thermal equilibrium it transfers 600 J of heat energy to gas B.

What is the initial thermal energy of B?

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  1. 23 October, 01:28
    0
    14657.32 J

    Explanation:

    Given Parameters;

    Number of moles mono atomic gas A, n 1 = 4.2 mol

    Number of moles mono atomic gas B, n 2 = 3.2mol

    Initial energy of gas A, K A = 9500 J

    Thermal energy given by gas A to gas B, Δ K = 600 J

    Gas constant R = 8.314 J / molK

    Let K B be the initial energy of gas B.

    Let T be the equilibrium temperature of the gas after mixing.

    Then we can write the energy of gas A after mixing as

    (3/2) n1RT = KA - ΔK

    ⟹ (3/2) x 4.2 x 8.314 x T = 9500 - 600

    T = (8900 x 3) / (2x4.2x8.314) = 382.32 K

    Energy of the gas B after mixing can be written as

    (3/2) n2RT = KB + ΔK

    ⟹ (3/2) x 3.2 x 8.314 x 382.32 = KB + 600

    ⟹ KB = [ (3/2) x 3.2 x 8.314 x 382.32] - 600

    ⟹ KB = 14657.32 J
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