Termodinamica Act 5.docx

Santiago Landeros Bautista U162058M0083 Ingeniería Industrial Termodinámica Actividad 5 19 de marzo de 2019

FORMULA ∆𝑉 = (𝑉𝑓 − 𝑉𝑖 )

𝑊 = −628215 𝐽

𝑝 = 1 𝑎𝑡𝑚

𝑄 = 1250 𝑘𝑐𝑎𝑙 = 5230000 𝐽

𝑉𝑓 = 18.2𝑚3 = 12000 𝑙𝑡𝑠

∆𝑈 = 𝑄 + 𝑊

𝑉𝑖 = 12𝑚3 = 18200 𝑙𝑡𝑠 ∆𝑉 = (18200 − 12000)𝑙𝑡𝑠 ∆𝑉 = (6200)𝑙𝑡𝑠 𝑊 = −𝑝(∆𝑉) 𝑊 = −1𝑎𝑡𝑚(6200)𝑙𝑡𝑠 𝑊 = −6200 𝑎𝑡𝑚 𝑙𝑡𝑠 101.325 𝑗 𝑊 = −6200 𝑎𝑡𝑚 𝑙𝑡( ) 1𝑙𝑡 𝑎𝑡𝑚 𝑊 = −628215 𝐽

∆𝑈 = 5230000 + ൫−628215 𝐽൯ = 4601785 𝐽

𝑉

𝑊 = −𝑅𝑛𝑇(𝑉2) 1

𝑊 = − ൬8.31

𝐽 7𝑚3 ∗ 𝐾൰ (2.60 𝑚𝑜𝑙)(290𝐾)𝐼𝑛 ቆ ቇ = −4343.08 𝐽 𝑚𝑜𝑙 3.5𝑚3

∆𝑈 = 0 ∆𝑈 = 𝑄 + 𝑊 𝑄 = −𝑊 = −൫−4343.08 𝐽൯ = 4343.08 𝐽

𝑁𝑜 𝑖𝑛𝑓𝑙𝑢𝑦𝑒 𝑒𝑛𝑒𝑟𝑔𝑖𝑎 𝑐𝑜𝑚𝑜 𝑐𝑎𝑙𝑜𝑟, 𝑑𝑎𝑑𝑜 𝑞𝑢𝑒 𝑒𝑠 𝑢𝑛 𝑝𝑟𝑜𝑐𝑒𝑠𝑜 𝑎𝑑𝑖𝑎𝑏𝑎𝑡𝑖𝑐𝑜 𝑄 = 0 𝑑𝑢 = 𝑊 + 𝑄 𝑄=0 𝑑𝑢 = 𝑊 𝑑𝑢 = 1850 𝐽 𝐸𝑠𝑒 𝑒𝑠 𝑒𝑙 𝑐𝑎𝑚𝑏𝑖𝑜 𝑑𝑒 𝑒𝑛𝑒𝑟𝑔𝑖𝑎 𝑙𝑖𝑏𝑟𝑒, 𝑒𝑛 𝑢𝑛𝑎 𝑐𝑜𝑚𝑝𝑟𝑒𝑠𝑖𝑜𝑛 𝑎𝑑𝑖𝑎𝑏𝑎𝑡𝑖𝑐𝑎 𝑙𝑎 𝑡𝑒𝑚𝑝𝑒𝑟𝑎𝑡𝑢𝑟𝑎 𝑠𝑢𝑒𝑚𝑝𝑟𝑒 𝑠𝑢𝑏𝑒

𝑃𝑞 = 1 𝑎𝑡𝑚 = 102325 𝑃𝑎 𝑉𝑄 = 3.5 𝐿𝑡 = 0.0035 𝑚3 𝑉𝑏 = 1.8 𝐿𝑡 = 0.0018𝑚3 𝑊 = 𝑃𝑎. 𝑉𝑎. 𝑙𝑛

𝑉𝑏 𝑉𝑎

𝑊 = 101325𝑃𝑎(0.0035𝑚3 )𝑙𝑛

0.0018𝑚3 0.0035𝑚3

𝑊 = 354.6375(−0.6649) = −235.82 𝐽