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termodinamika tolong jawab ya besok ujian ni! :((?

sejenis gas ideal dengan suhu awal 300 K mengalami proses isobarik pada tekanan 30 N/m2 . Dalam proses ini, kalor sebanyak 80 joule diberikan pada gas sehingga volumenya bertambah dari 2 m3 menjadi 6m3. hitunglah perubahan energi dalamnya! (pertanyaannya itu W nya nanti negatif atau positif kita buat? soalnya gak dikasih tau apakah gas melakukan usaha terhadap lingkungan atau sebaliknya) mohon dijawab master2 fisika!! :)

1 Answer

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  • Anonymous
    9 years ago
    Favorite Answer

    ΔQ = ΔU + W

    ΔQ = ΔU + PΔV

    80 = ΔU + 30(6 - 2)

    ΔU = -40 Joule

    periksa hasil di atas,

    untuk kondisi isobarik,

    V/T = V'/T'

    6/T = 2/300

    T = 900 K

    PV = nRT

    (30)(2) = n(8.314)(300)

    n = 0.0240558 mol

    ΔQ = ΔU + W

    ΔQ = ΔU + nRΔT

    80 = ΔU + (0.0240558)(8.314)(900 - 300)

    ΔU = -40 Joule

    ok cocok !

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