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Constant Volume Bomb Air-Water Analysis

1) All of the 1 lb of water introduced into the constant volume bomb evaporated, as the final partial pressure of water vapor (7.20243 psia) was less than the vapor pressure of water at 180°F (7.511 psia). 2) The volume of the bomb was calculated to be 43.80925 ft3 based on the ideal gas law and moles of dry air and water vapor present. 3) The final humidity of the air in the bomb was calculated to be 0.20812 lb of water per lb of air.

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Chrysan Hawa Nirwana
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184 views3 pages

Constant Volume Bomb Air-Water Analysis

1) All of the 1 lb of water introduced into the constant volume bomb evaporated, as the final partial pressure of water vapor (7.20243 psia) was less than the vapor pressure of water at 180°F (7.511 psia). 2) The volume of the bomb was calculated to be 43.80925 ft3 based on the ideal gas law and moles of dry air and water vapor present. 3) The final humidity of the air in the bomb was calculated to be 0.20812 lb of water per lb of air.

Uploaded by

Chrysan Hawa Nirwana
Copyright
© © All Rights Reserved
We take content rights seriously. If you suspect this is your content, claim it here.
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18.

7
A constant volume bomb contains air at 66 ℉ and 21.2 psia. One pound of liquid water is
introduced into the bomb. The bomb is then heated to a constant temperature of 180◦ F. After
equilibrium is reached, the pressure in the bomb is 33,0 psia. The vapor pressure of water at
180 ℉ is 7.51 psia.
a) Did all of the water evaporate?
b) Compute the volume of the bomb in cubic feet ?
c) Compute the humidity of the air the bomb at final conditions in pounds of water per
pound of air ?

 Di ketahui
 PADA KEADAAN AWAL
T = 66 = 66+459,67= 525,67◦R
udara
P = 21,2 psia

 PADA KEADAAN AKHIR (setelah ditambahkan H2O 1lb dan


di panaskan)
T=180℉ = 180+459,67=63,67 ◦R
P=33 psia
p*H2O( Vapor pressure of water at 80℉ ) = 7,511 𝑝𝑠𝑖𝑎

 Di tanya
a) Did all f the water evaporate ?
b) Compute the volume f the bomb i cubic feet ?
c) Cmpute the humidity of the air the bmb at final coditions in pounds
water per pounds f air ?

 Jawaban
 Step 1. mencari P H2O
Persamaan pada kodisi awal
Pair . Vair = nair . R . T
21,2 psia . Vair = nair . R . 525,67
Persamaa kodisi akhir
Pair . Vair = nair . R . T
( 33 – PH2O ) . Vair = nair . R . 634,63ºR
Dari kedua persamaan maka di dapatkan :
21,2 𝑝𝑠𝑖𝑎 . 𝑉𝑎𝑖𝑟 = 𝑛𝑎𝑖𝑟 . 𝑅 . 525,6 º𝑅
( 33 − 𝑃𝐻2𝑂 ). 𝑉𝑎𝑖𝑟 = 𝑛𝑎𝑖𝑟 . 𝑅. 639,67º𝑅

21,2 𝑝𝑠𝑖𝑎 525,67º𝑅


= (33 −𝑃𝐻2𝑂)
=
639,67º𝑅

= -525,67 . PH2O = - 3786,106

PH2O = 7,20243 psia

a) PH2O = 7,20243 psia


P*H2O = 7,511 psia
Iya semua air teruapkan karena di lihat dari perbedaan atara tekanan
parsial air dan tekanan uapnya , dari kedua tekanan tersebut dapat
diketahui bahwa tekanan parsial air lebih redah atau kurang dari tekanan
uapnya, sehingga seluruh air dapat teruapkan

 Step 2 mencari ndry air


1 𝑙𝑏
nH2O = = 0,055 lb.mol
18 𝑙𝑏/𝑚𝑜𝑙
Pdry air = Ptot – PH2O = 33 – 7,20243 = 25,776 psia
molar humidity :
PH2O 𝑛𝐻2𝑂
=
Pdry air 𝑛𝑑𝑟𝑦 𝑎𝑖𝑟

7,2024 𝑝𝑠𝑖𝑎 0,055 𝑙𝑏.𝑚𝑜𝑙


= =
25,7976 𝑝𝑠𝑖 𝑛𝑑𝑟𝑦 𝑎𝑖𝑟
0,055 𝑙𝑏.𝑚𝑜𝑙
= 0,28004 =
𝑛 𝑑𝑟𝑦 𝑎𝑖𝑟

= Ndry air = 0,16400 lb.mol

b) ntot = ndryair + nH2O = 0,16400 + 0,055 = 0,219


𝑛𝑡𝑜𝑡 . 𝑅 . 𝑇
V=
𝑃
𝑓𝑡3
(0,16400+0,055)𝑙𝑏𝑚𝑜𝑙 . 10,32 𝑝𝑠𝑖𝑎 . . 639,67◦𝑅
𝑙𝑏𝑚𝑜𝑙. 𝑅
=
33 𝑝𝑠𝑖𝑎
0,219. 10,32 𝑓𝑡3 . 639 ,67
=
33

1145,7053𝑓𝑡3
=
33

= 43,80925 ft3
𝑛𝐻2𝑂 𝑀𝑟 𝐻2𝑂
c) Humidity = X
𝑛 𝑑𝑟𝑦 𝑎𝑖𝑟 𝑀𝑟𝑑𝑟𝑦 𝑎𝑖𝑟

0,055 18 𝑙𝑏.𝑚𝑜𝑙
= 𝑋
0,16400 29 𝑙𝑏.𝑚𝑜𝑙
= 0,33536 x 0,6206

𝑙𝑏 𝐻2𝑂
= 0,20812
𝑙𝑏 𝑢𝑑𝑎𝑟𝑎

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