Boyle’s law yana daya daga cikin laws na gas wanda a makala ta da ta gabata na yi bayani akai. Kamar yadda na fada a baya, a yau kuma zamu je kai tsaye ne don duba daya daga cikin gas laws. Wannan law din shine boyle’s law. Wani lokaci akan kira boyle's laws da suna Boyle-Mariotte's law (musamman a kasar France). Wannan law din a karkashin binciken gwaji na gas law tana bayani ne akan yadda pressure na gas ya ke kokarin karewa a yayinda volume na gas din ya ke kokarin raguwa.
Sunan wannan law din ya samo asali ne daga sunan masanin kimiyya (chemist and physicist), Robert Boyle, wanda shine asalin wanda ya fara amfani da law din a shekarar 1662.
Ga bayanin boyle's law na wannan zamani a Turance kamar haka:
the absolute pressure exerted by a given mass of an ideal gas is inversely proportional to the volume it occupies if the temperature and amount of gas remain unchanged within a closed system.
Sannan kuma idan muka koma ainihin defination na law din a Turance shine kamar haka:
Boyle’s law states that volume (V) of a fixed mass of gas is inversely proportional to its pressure (P), provided the temperature remains constant.
Wannan shine law din wanda yanzu zan kawo yadda ya ke a karkashin formula wato yadda za a samu saukin aiki da shi wajen lissafi.
V α 1/P, yana nufin volume is inversely proportional to the pressure.
PV = K, wannan yana nufin pressure ayi multiplyin din shi da volume zai zama constant k, wanda P tana matsayin pressure na gas, V tana matsayin volume na gas,sai kuma k yana matsayin constant.
Wannan equation din yana bayani ne akan product na pressure da volume yana zama constant na mass of confined gas. Yana tsayawa a haka ne idan har temperature din ya zauna a matsayin constant. Idan aka kwatanta substance iri daya a karkashin condition guda biyu law din ana iya expressing dinta a matsayin, P1V1 = P2V2. Wannan formula tana nuna mana a lokacin da volume ya karu, pressure na gas yana raguwa in proportion. Haka kuma, idan volume na gas ya ragu, presssure na gas din yana karuwa.
Yanzu zamu dauki misalai mu ga yadda ake lissafi tare da wannan formula.
Misali na farko:
Amsar tambaya:
Wannan tambaya tana nuna mana gas wanda ya ke dauke da mass yana da pressure 650mmHg wanda ya mamaye volume 30cm3 . Tambayar da ake mana anan ita ce meye zai zama volume din gas din a 760 mmHg idan ya zama temperature din ya tsaya a constant,
Yanzu zamu fara fitar da data,
Data:
Initial gas pressure, P1 = 650mmHg
Final gas pressure, P2 = 760mmHg
Initial gas volume V1 = 30cm3
Daga P1V1 = P2V2, final gas volume, V2 = P1V2 / P2
V2 = 650mmHg × 30cm3 / 760mmHg = 25.66cm3
Misali na biyu:
Amsar tambaya:
Data:
Initial gas pressure, P1 = P0
Initial gas volume, V1 = V0
Final gas volume, V2 = 1 / 5V0
Yanzu zamu dauka daga P1 V1 = P2 V2, final gas pressure, P2 = P1 V1 /V2
P2 = P0V0 / 1/5V0 =P0V0 /1/5V0 = P0V0 × 5V0 = 5 P0
Application of Boyle’s Law to Gas Trapped in a Cylindrical Tube by a Column of Mercury
Zamu yi bayani ne akan application na boyle’s law akan gas wanda ya ke a cikin cylindrical tube daga column of mercury. Idan tube na uniform cross-section ya kulle a wani lungu yana dauke da fixed mass na gas ko na iska wanda an datse shi da column na mercury mai dauke da length X cm, pressure da ke jikin gas din ya danganta ne da plane of inclination na tube din kamar yadda yake a zanen fig 1.0
Fig 1.0
Pressure, p , na gas ya samu ne sakamakon sum na pressure exerted by X cm column of mercury da kuma atmospheric pressure,P .
Sakamakon haka za su zama, p = P + X (cm Hg).
2. Inverter with open end underneath
Fig 1.1
Gas pressure p ya samu ne sakamakon banbancin atmospheric pressure (P) da kuma pressure exerted by X cm column of mercury.
Sakamakon haka za su zama , p = P – X (cm Hg)
3. Horizontally
Fig 1.2
The gas pressure, p is due ONLY to the atmospheric pressure P, saboda idan tube din yana horizontal position, the column of mercury exert very negligible pressure akan gas. Saboda haka zai zama, p = P (cm Hg)
Generally, gas volume = length of column of gas (y) × cross-sectional area (A) of the tube.
Don haka gas volumes na fig a, b, c za su zama y1A, y2A, y3A cm3 .
Sun zama haka ne sakamakon cross sectional area na tube is uniform kuma duk daya ne a kowanne case, volume na gas ana daukar sa a matsayin length na column din gas. Shine ake nufi da volume V is proportional to length y.
saboda haka volume din gas na fig 1.0 shine y1 cm3
volume din gas na fig 1.1 shine y2cm3
volume din gas na fig 1.2 shine y3 cm3 .
Za a iya duba: Physics yadda ake lissafin screw
Boyle’s law (P1V1 = P2V2 ) can be applied in deriving a relationship between fig 1.0 and 1.1 yanzu zaku ga ta yadda ake fitar da formula wadda ke dauke da dangan takar dake tsakanin fig 1.0 da 1.1 a zanen da zai gabata da kuma ta hanyar nuna su a lissafi.
4. Fig 1.3
Daga zanen fig 1.3a
Initial or vertical volume, V1 = y1A
Daga fig 1.3 b
Final or inverted volume, V2 = y2A
Yanzu zamu yi substituting dinsu a cikin Boyle’s law P1V1 = P2V2 shine zai bamu
(P+X )y1A = ( P-X )y2A
Yanzu zamu yi dividing both side by A, shine zai bamu
(P+X)y1 = (P-X )y2
Tunda mun yi deriving wasu formula da ake amfani dasu wajan lissafin boyle’s law yanzu zamu dauko wata tambaya cikin tambayoyin da ake ma daliban sakandire ta karshen karatunsu wato a jarabawa, kamarsu WAEC mu amsa ta tare da amfani da wadannan fomulolin da muka koya a baya.
Misali na uku:
A uniform capillary tube closed at one end contained dry air trapped by a thread of mercury 8.5 × 10-2m long. When the tube was held horizontally, the length of the air column was 5.0 × 10-2m, when it was held vertically with the closed end downwards, the length was 4.5 × 10-2m . Determine the value of the atmospheric pressure. [ g = 10m/s-2, density of mercury = 1.36 × 104kgm-3]. WEAC 2004
Amsar tambaya:
Abinda ya kamata mu fara yi shine mu fitar da zane wanda a kai mana bayaninsa a wannan tambayar hakan zai sa musamu saukin lissafi mu kuma gane mai muke so mu nemo.
Fig 1.4
Yanzu tunda mun ga yadda suke a zanan, zamu wuce kai tsaye mu fitar da data.
At horizontal position, gas pressure is due only to atmospheric pressure, P.
Daga fig 1.4a, above, Horizontal gas pressure, P 1 = P
Horizontal gas volume, V1 = 0.05A
Where A is cross sectional area.
At vertical position, gas pressure is due to atmospheric pressure + pressure due to mercury column.
From fig 1.4b above, vertical gas pressure, P2 = P + 0.085
Vertical gas volume, V2 = 0.045A
Yanzu zamu sasu cikin boyle’s law
Substituting into boyle’s law, P1V1 = P2V2 , we obtain
P × 0.05A = (P + 0.085) × 0.045A
Divide both sides by A
P × 0.05 = (P + 0.085) × 0.045
0.05P = 0.045P + 0.003825
0.05P – 0.045P = 0.003825
0.005P = 0.003825
P = 0.003825/0.005 = 0.765mHg
The atmospheric pressure (0.765mHg) so obtained is in length of mercury and has to be converted to N/m2 or Pa using the pressure formula, P = ρgh.
Given values: density of mercury, ρ = 1.36 × 104kgm-3; g = 10m/s2
Calculated value: height of mercury, h = 0.765mHg
Therefore, atmospheric pressure, P = ρgh
= 1.36 × 104kgm-3 × 10 × 0.765
104040Nm-2 or 1.04 × 105Pa
Wadannan sune kadan daga cikin misalan lissafin boyle’s law,yana da kyau dalibi ya kara bincike sosai da kuma amsa wasu tambayoyin domin kwarewa a wannan topic. Allah Ya sa mu dace, amin.
Mai karatu na iya duba: lissafin weight of a body akan lift da sauransu.
Hakkin mallakar hoto (photo credit): googlegalaxysecience
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