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A) Calculate the work done on or by the helium gas in joules (J).  B) What is the change in the helium’s internal energy in kilojoules (kJ)?

A total of 22.6 kJ of heat energy is added to a 6.05L sample of helium at 0.986atm. The gas is allowed to expand against a fixed external pressure to a volume of 24.5L.

1 Answer

  • 1 month ago


    (24.5 - 6.05)L x (0.986 atm x (101.325 kPa/atm)) = 1843 J = -1.843 kJ

    (The last minus sign shows that the work was done by the He on the surroundings.)


    ΔE = +22.6 kJ - 1.843 kJ = +20.8 kJ

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