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Bölüm 2: Termodinamiğin 1. KanunuQ = ∆U+ WQ = −10210.1−0.015Q = −10210. 1015 j (Çevreden sisteme ısı verilmiştir)Özet olarak:P (Pa)3∆U∆HQW31313131= −10210j= −17016.9j= −10210.1015j= −0.015j∆U∆HQW23232323= 6807.42 j= 11345.7 j= 6807.42 j= 01.013×10 52.026×10 5 1 2∆U12= 3402.71j∆H12= 5671.19 jQ12= 5671.83 jW12= 2269.12 j0.0224 0.0448V (m 3 )18