Chemistry

A gas is confined to a cylinder under constant atmospheric pressure, as illustrated in the following figure. When the gas undergoes a particular chemical reaction, it absorbs 827 J of heat from its surroundings and has 0.66 kJ of P−V work done on it by its surroundings. Calculate Delta H and Delta E.

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  1. delta H = q = 827 J and since it absorbs heat the sign is +.
    delta E = q + work
    work = PdV = 0.66 kJ
    Since work is done ON IT, the sign of work is +.
    Substitute and solve.

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  2. 827.66

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  3. You just convert the absorption of heat into kJ
    So 827 J of heat absorbed = .827 kJ

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  4. ^ that is for delta H

    Delta E is going to be kJ of heat absorbed plus kJ of work
    So in this case it is .827 kJ + .66 kJ = 1.49 kJ

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