Misalan Tambayoyi 20 na Thermodynamics
Tsarin Thermodynamic
1. A cikin thermodynamics, iskar gas mai kyau tana fuskantar tsarin isothermal idan…
A. canje-canje a yanayin iskar gas, zafin jiki koyaushe yana nan daidai
B. duk kwayoyin halitta suna motsawa a cikin sauri daban-daban
C. Duk yanayin iskar gas koyaushe yana canza yanayin zafi
D. a yanayin zafi mai yawa, saurin ƙwayoyin ya fi girma
E. matsin lamba da ƙarar iskar gas ba sa canzawa
Tattaunawa
Isothermal ko isothermic yana nufin yanayin zafi mai ɗorewa.
Amsar da ta dace ita ce A.
2. Duk iskar gas mai kyau suna fuskantar tsari na isochoric don haka...
A. Duk kwayoyin halitta suna da irin wannan gudu
B. a yanayin zafi mai yawa matsakaicin saurin ƙwayoyin halitta ya fi girma
C. matsin iskar gas ya kasance iri ɗaya
D. iskar gas za ta yi aiki
E. ba shi da makamashin ciki
Tattaunawa
Isochoric = ƙarar da ba ta canzawa.
Amsar da ta dace ita ce B.
3. Bayanin da ya dace game da hanyoyin thermodynamic shine…
A. Isobaric tsari ne na canza iskar gas a matsin lamba akai-akai.
B. Isobaric tsari ne na canza iskar gas a yanayin zafi mai ɗorewa.
C. Isochoric tsari ne na canza iskar gas a matsin lamba akai-akai.
D. isotherm shine tsarin canza iskar gas a daidai girman da ake buƙata.
E. Isochoric tsari ne na canza iskar gas a yanayin zafi mai ɗorewa.
Tattaunawa
Isobaric = matsin lamba akai-akai
Isochoric = ƙarar da ba ta canzawa
Isothermal = yanayin zafi mai ɗorewa
Amsar da ta dace ita ce A.
Dokar Farko ta Thermodynamics
4. Daga jadawalin PV da ke ƙasa, adadin aikin iskar gas a cikin matakai na I da II yana daidai gwargwado kamar...
A. 4: 3
B. 3: 4
C. 2: 3
D. 1: 2
E. 1: 1
Tattaunawa
An san cewa:
Tsarin 1:
Matsi (P) = 20 N/m 2
Ƙarar farko (V1 ) = lita 10 = 10 dm 3 = 10 x 10 -3 m 3
Ƙarar ƙarshe (V2 ) = lita 40 = 40 dm 3 = 40 x 10 -3 m 3
Tsarin 2:
Matsi (P) = 15 N/m 2
Ƙarar farko (V1 ) = lita 20 = 20 dm 3 = 20 x 10 -3 m 3
Ƙarar ƙarshe (V2 ) = lita 60 = 60 dm 3 = 60 x 10 -3 m 3
An tambaya:
Amsa:
Aikin iskar gas na aiki I:
W = P ΔV = P (V2 –V1 ) = (20)(40-10)(10 -3 m3 ) = (20)(30)(10 -3 m3 ) = (600)(10 -3 m3 ) = 0,6 m3
Aikin iskar gas a cikin tsari na II:
W = P ΔV = P (V2 –V1 ) = (15)(60-20)(10 -3 m3 ) = (15)(40)(10 -3 m3 ) = (600)(10 -3 m3 ) = 0,6 m3
Kwatanta aikin iskar gas a cikin matakai I da II:
0,6 m 3 : 0,6 m 3
1: 1
Amsar da ta dace ita ce E.
5. Daga jadawalin P – V, adadin aikin iskar gas a cikin matakai I da II yana daidai gwargwado kamar...
A. 4: 3
B. 3: 4
C. 2: 3
D. 1: 2
E. 1: 1
Tattaunawa
An san cewa:
Tsarin I:
Matsi (P) = 20 Pascal
Juzu'i na 1 (V 1 ) = lita 10
Juzu'i na 2 (V 2 ) = lita 40
Tsari na II:
Matsi (P) = 15 Pascal
Juzu'i na 1 (V 1 ) = lita 20
Juzu'i na 2 (V 2 ) = lita 60
Tambaya: Kwatanta aikin iskar gas a cikin matakai I da II
Amsa:
Tsarin thermodynamic bisa ga jadawalin da ke sama tsari ne na isobaric (matsi mai ɗorewa).
Aikin da iskar gas ke yi a cikin tsari I:
W = P (V 2 – V 1 )
W = (20)(40 – 10)
W = (20)(30)
W=600
Aikin da iskar gas ke yi a cikin tsari na II:
W = P (V 2 – V 1 )
W = (15)(60 – 20)
W = (15)(40)
W=600
Kwatanta aikin iskar gas a cikin matakai I da II:
600: 600
1: 1
Amsar da ta dace ita ce E.
6. Ana dumama iskar helium bisa ga jadawalin PV da ke ƙasa. Aikin da iskar helium ke yi a cikin tsarin AB shine...
A. 15 wasa 
B. 10 joules
C. 8 joules
D. 4 joules
E. 2 joules
Tattaunawa
An sani cewa :
Matsi (P) = 2 x 10 5 N/m 2 = 2 x 10 5 Pascal
Ƙarar farko (V1 ) = 5 cm 3 = 5 x 10 -6 m 3
Ƙarar ƙarshe (V2 ) = 15 cm 3 = 15 x 10 -6 m 3
Tambaya : Aikin da iskar gas ta yi a tsarin AB
Amsa :
W = ∆P ∆V
W = P (V 2 – V 1 )
W = (2 x 10 5 )(15 x 10 -6 – 5 x 10 -6 )
W = (2 x 10 5 )(10 x 10 -6 ) = (2 x 10 5 )(1 x 10 -5 )
W = Joule 2
Amsar da ta dace ita ce E.
7. Kalli hoton! Iskar gas mai kyau tana fuskantar tsarin canza matsin lamba (P) zuwa girma (V). Aikin da iskar gas ke yi a wannan tsari shine…
A. Joule 20
B. 15 Joule
C. 10 Joule
D. 5 Joule
E. 4 Joule
Tattaunawa
An sani cewa :
Matsi na farko (P1 ) = 4 Pa = 4 N/ m2
Matsi na ƙarshe (P2 ) = 6 Pa = 6 N/ m2
Ƙarar farko (V1 ) = 2 m3
Ƙarar ƙarshe (V2 ) = 4 m3
Ana nema : Aikin da gas (W) ya yi
Amsa :
Aikin da gas ke yi = yankin da ke ƙarƙashin lanƙwasa ab.
W = yankin alwatika + yankin murabba'i mai kusurwa huɗu
W = ½ (6-4)(4-2) + 4(4-2)
W = ½ (2)(2) + 4(2)
W = 2 + 8
W = Joule 10
Amsar da ta dace ita ce E.
8. Iskar gas mai kyau tana fuskantar tsari mai rufewa A → B → C → A. A cikin zagayowar, iskar gas tana aiki gwargwadon….
A. -2,0 x 103 J
B. − 5,5 x 10 3 J
C. − 8,0 x 10 5 J
D. 2,0 x 10 6 J
E. 4,0 x 10 6 J
Tattaunawa
Aiki (W) = yankin lanƙwasa (yankin alwatika a cikin layin da aka yiwa alama da kibiya).
W = ½ (20-10)(6 x 10 5 – 2 x 10 5 )
W = ½ (10)(4 x 10 5 )
W = (5)(4 x 10 5 )
W = 20 x 10 5 = 2 x 10 Joules 6
Amsar da ta dace ita ce D.
Injin Zafi
9. Inji yana shan Joules 2.000 na zafi daga wani ma'ajiyar zafi mai zafi sosai kuma yana ƙin Joules 1.200 zuwa wani ma'ajiyar zafi mai ƙarancin zafi. Ingancin injin shine ....
A. 80%
B. 75%
C. 60%
D. 50%
E. 40%
Tattaunawa
An san cewa:
Zafi da aka sha (QH ) = Joules 2000
Zafi da aka saki (QL ) = Joules 1200
Aikin da injin ya samar (W) = 2000 – 1200 = Joules 800
Tambaya : Ingancin injin zafi (e)
Amsa :
Tsarin da ke tabbatar da ingancin injin zafi:
e = W / Q H = 800/2000 = 0,4 x 100% = 40%
Amsar da ta dace ita ce E.
Injin mota
10. Ingancin injin Carnot wanda zagayowarsa ke shan zafi a zafin jiki na 960 K kuma yana ƙin zafi a zafin jiki na 576 K shine...
A. 40%
B. 50%
C. 56%
D. 60%
E. 80%
Tattaunawa
An sani cewa :
Zafin jiki mai yawa (TH ) = 960 K
Ƙananan zafin jiki (TL ) = 576 K
Tambaya : Ingancin injin Carnot (e)
Amsa :
Ingancin injin Carnot shine 0,4 x 100% = 40%
Amsar da ta dace ita ce A.
11. A cikin jadawalin PV na injin Carnot da ke ƙasa, W = Joules 6.000. Adadin zafi da injin ke fitarwa a kowace zagaye shine...
A. 2.250 joules
B. 3.000 joules
C. 3.750 joules
D. 6.000 joules
E. 9.600 joules
Tattaunawa
An san cewa:
Aiki (W) = Joules 6000
Zafin jiki mai yawa (TH ) = 800 Kelvin
Ƙananan zafin jiki (T L ) = 300 Kelvin
An tambaya: Q
Amsa:
Ingancin injin zafi mai kyau (injin Carnot):
Zafin da injin Carnot ke sha:
W = e Q1
6000 = (0,625) Q1
Q1 = 6000 / 0,625
Q1 = 9600
Zafin da injin Carnot ke fitarwa:
Q2 = Q1 – W
Q2 = 9600 – 6000
Q2 = Joule 3600
Babu amsar da ta dace.
12. Injin Carnot mai inganci na kashi 40% yana amfani da wurin ajiyar ruwa mai zafi a zafin jiki na 727°C. Kayyade zafin wurin ajiyar ruwan sanyi!
A. 327°C
B. 357°C
C. 400°C
D. 600°C
E. 627°C
Tattaunawa
An san cewa:
Inganci (e) = 40% = 40/100 = 0,4
Zafin jiki mai yawa (TH ) = 727 o C + 273 = 1000 K
Tambaya: Ƙayyade zafin wurin ajiyar ruwan sanyi
Amsa:
Zafin ma'ajiyar shine 600–273 = 327 oC
Amsar da ta dace ita ce A.
13. Jadawalin P-V na injin Carnot yayi kama da hoton da ke ƙasa! Idan injin ya sha zafi 800 J, to aikin da aka yi shine…
A. 105,5 J
B. 252,6 J
C. 336,6 J
D. 466,7 J
E. 636,7 J
Tattaunawa
An san cewa:
Zafin jiki mai yawa (TH ) = 600 Kelvin
Ƙananan zafin jiki (T L ) = 250 Kelvin
Zafi da aka sha (Q1 ) = Joules 800
An tambaya: Ƙoƙari (W)
Amsa:
Ingancin injin zafi mai kyau (injin Carnot):
Kokarin da aka yi sune:
W = e Q1
W = (7/12)(Joules 800)
W = Joule 466,7
Amsar da ta dace ita ce D.
14. Injin Carnot yana aiki a babban zafin jiki na 600 K, don samar da aikin injiniya. Idan injin ya sha zafi mai girman 600 J a ƙaramin zafin jiki na 400 K, to aikin da aka samar shine….
A. 120 J
B. 124 J
C. 135 J
D. 148 J
E. 200 J
Tattaunawa
An sani cewa :
Ƙananan zafin jiki (TL ) = 400 K
Zafin jiki mai yawa (TH ) = 600 K
Zafi da aka sha (Q1 ) = Joules 600
Ana Neman Aiki : Aikin da injin Carnot (W) ya samar
Amsa :
Ingancin injin zafi mai kyau (injin Carnot):
Aikin da injin Carnot ya yi:
W = e Q1
W = (1/3)(600) = Joule 200
Amsar da ta dace ita ce E.
Ka'idar Kinetic na Gases da Thermodynamics
15. A cikin tanki akwai iskar gas mai kyau wacce take da lita 4, zafin jiki na 27 oC da matsin lamba na atm 3 (1 atm = 10 5 Nm -2 ). Iskar tana fuskantar tsarin dumamawa a matsin lamba akai-akai zuwa zafin jiki na 87 oC . Ƙarfin zafin iskar gas shine 9 JK -1 . Ƙarar ƙarshe ta iskar gas da canjin makamashin ciki na iskar gas ɗin bi da bi...
A. lita 4,2, ΔU = Joules 200
B. Lita 4,4, ΔU = Joules 240
C. lita 4,6, ΔU = Joules 280
D. lita 4,8, ΔU = Joules 300
E. lita 4,8, ΔU = Joules 360
Tattaunawa
Tsarin isobaric (matsi mai ɗorewa)
An san cewa:
Girman farko na iskar gas mai kyau (V1 ) = lita 4
Zafin farko na iskar gas mai kyau (T 1 ) = 27 o C + 273 = 300 K
Zafin ƙarshe na iskar gas mai kyau (T2 ) = 87 o C + 273 = 360 K
Matsin iskar gas mai kyau (P) = 3 atm = 3 x 10 5 Nm -2
Ƙarfin zafi na iskar gas (C) = 9 JK -1
Tambaya: Ƙarar gas ta ƙarshe (V2 ) da Canjin makamashin ciki na iskar gas (ΔU)
Amsa:
Lissafa ƙarar ƙarshe ta amfani da dabarar dokar Charles (tsarin isobaric ko matsin lamba mai ɗorewa):
Canjin Girma:
Lita 1 = 0,001 m3
Ƙarar farko (V1 ) = 4 (0,001 m3 ) = 0,004 m3
Ƙarar ƙarshe (V2 ) = 4,8 (0,001 m3 ) = 0,0048 m3
Canjin girma ( ΔV) = V 2 – V 1 = 0,0048 m 3 – 0,004 m 3 = 0,008 m 3.
Canje-canjen Zafin Jiki:
Canjin zafin jiki (ΔT) = T 2 – T 1 = 360 K – 300 K = 60 K
Lissafa canjin makamashin ciki (ΔU) na iskar gas mai kyau ta amfani da dabarar Dokar Thermodynamics ta Farko:
ΔU = Q-W
Bayani: ΔU = canjin kuzarin ciki, Q = zafi, W = aiki.
Lissafa aikin (W) a matsin lamba akai-akai:
W = P ΔV = (3 x 10 5 )( 0,0008) = (3 x 10 1 )( 8) = (30)(8) = Joules 240
Lissafin zafi (Q) ta amfani da dabarar ƙarfin zafi (C):
C = Q / ΔT
Q = (C) (ΔT) = (9) (60) = 540 Joules
Lissafa canjin makamashin ciki:
ΔU = Q – W = Joules 540 – Joules 240 = Joules 300.
Amsar da ta dace ita ce D.
16. A cikin tanki akwai lita 6 na iskar gas mai kyau tare da matsin lamba na atm 2 (atm 1 = 10 5 Nm -2 ), zafin jiki na 27 oC . Ana dumama iskar gas ɗin zuwa zafin jiki na 77 oC a matsin lamba akai-akai. Idan ƙarfin zafin gas ɗin shine 5 JK -1 , ƙarar ƙarshe da canjin makamashin ciki na gas ɗin bi da bi ne….
A. lita 8; ΔU = Joules 250
B. Lita 8; ΔU = Joules 200
C. lita 7; ΔU = Joules 100
D. lita 7; ΔU = Joules 50
E. lita 7; ΔU = Joules 20
Tattaunawa
Tsarin isobaric (matsi mai ɗorewa)
An san cewa:
Girman farko na iskar gas mai kyau (V1 ) = lita 6
Zafin farko na iskar gas mai kyau (T 1 ) = 27 o C + 273 = 300 K
Zafin ƙarshe na iskar gas mai kyau (T2 ) = 77 o C + 273 = 350 K
Matsin iskar gas mai kyau (P) = 2 atm = 2 x 10 5 Nm -2
Ƙarfin zafi na iskar gas (C) = 5 JK -1
Tambaya: Ƙarar gas ta ƙarshe (V2 ) da Canjin makamashin ciki na iskar gas (ΔU)
Amsa:
Lissafa ƙarar ƙarshe ta amfani da dabarar dokar Charles (tsarin isobaric ko matsin lamba mai ɗorewa):
Canjin Girma:
Lita 1 = 0,001 m3
Ƙarar farko (V1 ) = 6 (0,001 m3 ) = 0,006 m3
Ƙarar ƙarshe (V2 ) = 7 (0,001 m3 ) = 0,007 m3
Canjin girma ( ΔV) = V 2 – V 1 = 0,007 m 3 – 0,006 m 3 = 0,001 m 3.
Canje-canjen Zafin Jiki:
Canjin zafin jiki (ΔT) = T 2 – T 1 = 350 K – 300 K = 50 K
Lissafa canjin makamashin ciki (ΔU) na iskar gas mai kyau ta amfani da dabarar Dokar Thermodynamics ta Farko:
ΔU = Q-W
Bayani: ΔU = canjin kuzarin ciki, Q = zafi, W = aiki.
Lissafa aikin (W) a matsin lamba akai-akai:
W = P ΔV = (2 x 10 5 )( 0,001) = (2 x 10 2 )(1 ) = (200)(1) = Joules 200
Lissafin zafi (Q) ta amfani da dabarar ƙarfin zafi (C):
C = Q / ΔT
Q = (C) (ΔT) = (5) (50) = 250 Joules
Lissafa canjin makamashin ciki:
ΔU = Q – W = Joules 250 – Joules 200 = Joules 50.
Amsar da ta dace ita ce D.
Tsarin Thermodynamic
17. Duba jadawalin da ke gefe da kuma waɗannan maganganun!
(1Tsarin AB shine isobaric kuma W = P (V)2 - V1)
(2) Tsarin BC adiabatic ne, kuma ΔU = Q
(3) Tsarin BC shine isochoric, kuma ΔU = Q
(4) Tsarin CA yana da isothermal, kuma tsarin yana shan zafi.
Bayanin da ya dace shine…
A. (1) da (3)
B. (1) da (4)
C. (2) da (3)
D. (2) da (4)
E. (3) da (4)
Tattaunawa
(1 ) Tsarin AB shine isobaric kuma W = P ( V2 - V1 )
Isobaric = matsin lamba mai dorewa. Ana nuna matsin lamba mai dorewa ta hanyar layi madaidaiciya tare da P1 daga A zuwa B. A cikin wannan tsari, matsin lambar yana ci gaba da kasancewa akai-akai amma akwai canji a girma inda ƙarar ke ƙaruwa. Idan ƙarar ta ƙaru, tsarin yana aiki akan muhalli, inda adadin aikin (W) sakamakon samfurin matsin lamba (P) ne da canjin girma (ΔU).
(3) Tsarin BC shine isochoric, kuma ΔU = Q
Isochoric = ƙara mai ɗorewa. Ana nuna ƙara mai ɗorewa ta hanyar layi madaidaiciya tare da V 2 daga B zuwa C. A cikin wannan tsari, ƙarar ta kasance mai ɗorewa amma akwai canji a matsin lamba inda matsin ya ƙaru. Ƙarar ta kasance mai ɗorewa don haka babu wani aiki da aka yi, inda W = 0. Tsarin Dokar Farko ta Thermodynamics shine ΔU = Q - W, inda ΔU = canji a cikin kuzarin ciki, Q = zafi da W = aiki. Idan babu aiki to W = 0, don haka ΔU = Q.
Amsar da ta dace ita ce A.
18. Kalli jadawalin zagayowar thermodynamic mai zuwa!
Daga wannan bayanin:
(1) A - B = isochoric, aiki ya dogara da canje-canje a yanayin zafi
(2) B – C = adiabatic, babu canjin zafi
(3) B - C = isobaric, aiki ya dogara da canje-canje a cikin girma
(4) C - A = tsarin yana faruwa ne da ƙoƙari daga wajen tsarin
Gaskiyar magana ita ce…
A. (1) da (2)
B. (1) da (3)
C. (2) da (3)
D. (2) da (4)
E. (3) da (4)
Tattaunawa
(1) A - B = isochoric, aiki ya dogara da canje-canje a yanayin zafi
Eh, wannan tsari yana da isochoric (ƙarfin da ba ya canzawa). Duk da haka, ƙarar tana nan daram, don haka babu wani aiki da aka yi. Aiki yana faruwa ne idan aka sami canji a ƙarar. Saboda haka, wannan magana ba daidai ba ce.
(3) B - C = isobaric, aiki ya dogara da canje-canje a cikin girma
Isobaric = matsin lamba mai ɗorewa. Ana nuna matsin lamba mai ɗorewa ta hanyar layi madaidaiciya daga B zuwa C. A cikin wannan tsari, akwai canji a cikin girman tsarin, inda girman tsarin ke ƙaruwa. Ƙara girman tsarin yana nufin tsarin yana aiki akan muhalli.
(4) C - A = tsarin yana faruwa ne da ƙoƙari daga wajen tsarin
A cikin wannan tsari, ƙarar tsarin yana raguwa kuma matsin lamba na tsarin yana raguwa. Rage girman tsarin yana nufin cewa muhalli yana aiki akan tsarin. A wata ma'anar, ana yin aiki a waje.
Amsar da ta dace ita ce E.
Dokokin Thermodynamics
19. Kalli hoton! Iskar gas mai kyau tana fuskantar tsarin canza matsin lamba (P) zuwa girma (V). Aikin da iskar gas ke yi a wannan tsari shine…
A. Joule 20
B. 15 Joule
C. 10 Joule
D. 5 Joule
E. 4 Joule
Tattaunawa
An sani :
Matsi na farko (P)1) = 4 Pa = 4 N/m2
Matsi na ƙarshe (P)2) = 6 Pa = 6 N/m2
Ƙaramin farko (V)1= 2m3
Ƙarar ƙarshe (V)2= 4m3
An tambaya Aikin da gas (W) ya yi
Jawab :
Aikin da gas ke yi = yankin da ke ƙarƙashin lanƙwasa ab.
W = yankin alwatika + yankin murabba'i mai kusurwa huɗu
W = ½ (6-4)(4-2) + 4(4-2)
W = ½ (2)(2) + 4(2)
W = 2 + 8
W = Joule 10
Amsar da ta dace ita ce E.
20. Iskar gas mai kyau tana fuskantar tsari mai rufewa A → B → C → A. A cikin zagayowar, iskar gas tana aiki gwargwadon….
A. −2,0 x 103 J
B. −5,5 x 103 J
C. −8,0 x 105 J
D. 2,0 x 106 J
E. 4,0 x 106 J
Tattaunawa
Aiki (W) = yankin lanƙwasa (yankin alwatika a cikin layin da aka yiwa alama da kibiya).
W = ½ (20-10)(6 x 10)5 - 2 x 105)
W = ½ (10)(4 x 105)
W = (5)(4 x 105)
W = 20 x 105 = 2 x106 Joule
Amsar da ta dace ita ce D.
Tushen tambaya:
Tambayoyin Nazarin Fizik na Ƙasa ga Makarantar Sakandare ta Babbar Sakandare/Makarantar Sakandare ta Sana'a
