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1.2. Topshiriqni bajarish uchun misol
Misol. Quyidagi parametrlar asosida (xonani uzunligi l=4,0 m; xonani kengligi n=2,0 m; xonani balandligi h=3,0 m. yonilgI ogirligi m=12 kg; yonish koefitsienti k=0,8; Ψ1=0,1 to‘liq yonib ulgurmagan (CO hosil qiluvchi) uglerod miqdoriga javob beruvchi koefitsient; Ψ2=0,15 ikkilamchi jaryonda CO hosil bo‘lishiga javob beruvchi koefitsient. T1=400C=313K; P=780mm.rt.st.) yog‘och, ko‘mir yoki boshqa turdagi yonilg‘ilarni yopiq xonada toliq yonishida xonananing qaysi balnadligida is gazini toplanishini va is gazi qanday xajimni egallashini toping. Sodda qilib aytganda, uglerod oksidi yuqori qismida joylashgan va boshqa gazlar bilan aralashmaydi deb taxmin qilamiz.
Hisoblash: Barcha yoqilgan yoqilg'ini toza uglerod deb taxmin qilamiz. Uning miqdori yonish koeffitsienti bilan yoqilg'i massasi mahsuloti bilan aniqlanadi:
(1.5)
yoki
Yoqilgini yonishi parallel holda ikkita jarayonda boradi:
(1.6)
(1.7)
Karbonad angidrid bir qismi choglangan komir bilan ikkilamchi reaksiyaga kirishadi:
(1.8)
(1.6) da keltirilgan reaksiyada qatnashadigan uglerodning massasi quyidagiga teng:
(1.9)
yoki
(1.7) da keltirilgan reaksiyada qatnashadigan uglerodning massasi quyidagiga teng:
(1.10)
yoki
CO ni hosil qiluvchi uglerodning umumiy massasi quyidagiga teng:
(1.11)
yoki
Osonlashtirish maqsadida is gazini hosil bolish jarayonini (1.7) dagi reaksiya boyicha olib borilmoqda deb hisoblaymiz. Kimyoviy reaktsiyaga jalb qilingan massalarning nisbati asosida (1.1-vazifani echish uchun tushuntirishlarga qarang), hosil bolgan uglerod oksidining massasini topamiz.
(1.12)
yoki .
(CO ni molekular ogirligini uglerod va kislorod atom ogirligi yigindisidek topamiz; (1.7) tenglamadagi CO va C dan oldingi koeffitsientlar bir-birini bekor qiladi).
Normal sharoitlarda ushbu miqdordagi uglerod oksidi olinadigan hajm qoyidagich hisoblanadi:
yoki 4,480 m3.
(CO ning bir mol massasi - 0,028 kg; normal sharoitda bir mol gaz egallagan hajmi - 22,4 litr, 1.1 vazifani echish uchun tushuntirishlarga qarang).
Birlashtirilgan gaz qonunining tenglamasiga binoan biz T = 313K darajasida uglerod oksidining haqiqiy hajmini topamiz:
(1.13)
Bu yerda
Xonaning maydoni
Uglerod oksidi bilan to'ldirilgan zonaning balandligini aniqlaymiz:
Korinib turibtiki xonada is gazi miqdori meyordan yuqori yoki
Javob: uglerod oksidi bilan to'ldirilgan zona 2,375 metrdan yuqori.
1.2-jadval
Vazifani bajarish uchun variantlar
Вариант рақамларини паспортингизни охирги рақами бўйича оласиз (АА2568794)
№ variant
|
m, kg
|
T1, 0C
|
P1, mm.rt.st.
|
K
|
Ψ1
|
Ψ2
|
l, m
|
n, m
|
h, m
|
|
15
|
42
|
780
|
0.75
|
0.1
|
0.15
|
2
|
4
|
2
|
|
25
|
46
|
784
|
0.83
|
0.18
|
0.17
|
2.5
|
5
|
3.7
|
|
17
|
50
|
786
|
0.82
|
0.19
|
0.18
|
8
|
3
|
2.75
|
|
24
|
54
|
785
|
0.76
|
0.17
|
0.19
|
3
|
6
|
2.7
|
|
19
|
40
|
788
|
0.79
|
0.2
|
0.14
|
3
|
3
|
3
|
|
31
|
58
|
787
|
0.77
|
0.3
|
0.12
|
2
|
4
|
2
|
|
26
|
52
|
783
|
0.78
|
0.21
|
0.13
|
2.5
|
5
|
3.7
|
|
10
|
48
|
782
|
0.84
|
0.16
|
0.11
|
8
|
3
|
2.75
|
|
21
|
44
|
789
|
0.85
|
0.14
|
0.1
|
3
|
6
|
2.7
|
|
37
|
56
|
781
|
0.8
|
0.15
|
0.2
|
3
|
3
|
3
|
1.2. Topshiriqni bajarish uchun
Javoblar: (bu yerda oz variantingizni berilishini yozib kiyin hisoblaysiz)
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