Question

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Answer 1

1.

PV = nRT

in conditii normale (0oC si P=1 atm) avem 448 L N2

n = V/Vm = 448/22,4 = 20 moli N2

P = ? atm , V= 8 L , T= 273+40 = 313 K

=> P = nRT/V = 20x0,082x313/8 = 64,165 atm

2.

n1 kmoli     n2 kmoli     138 kg

   3H2   +   N2     -->   2NH3

    3             1                2x17

=> n1 = 3x138/2x17 = 12,18 kmoli H2 = 12180 moli

=> n2 = 1x138/2x17 = 4,06 kmoli N2 = 4060 moli

=> n.amestec = n1+n2 = 12180+4060 = 16240 moli amestec

PV = nRT

P= 4 atm , T= 273+117 = 400 K

=> V = nRT/P

= 16240x0,082x400/4 = 133168 L = 133,168 m3

pt. NH3

138 kg/17 = 8,12 kmoli = 8120 moli

PV = nRT

P= 4 atm , T= 273+117 = 400 K

=> V = nRT/P

= 8120x0,082x400/4 = 66560 L = 66,56 m3 conditii date

V/Vm = m/M => V = Vmxm/M

= 22,4x138/17 = 181,84 m3 conditii normale

=> V.conditii.normale/V.conditii.date = 181,84/66,56 = 2,73 ori