Makalu

Bayanai game da Charle's Law

  • Charle’s law na daya daga cikin laws na gas wanda a makala ta da ta gabata na yi bayani akai. Yau kuma zan yi bayani ne, kamar yadda na ambata a baya, akan daya daga cikin gas law din wato charle’s law. Wannan law din ya samo sunansa ne daga wani masanin kimiyya mai suna, Jacques Charles.

    Ga bayanin Charle’s law a Turance kamar haka:

    The volume of a given mass of an ideal gas is directly proportional to its temperature on the absolute temperature scale (in Kelvin) if pressure and the amount of gas remain constant; that is, the volume of the gas increases or decreases by the same factor as its temperature.

    Sannan kuma daliban sakandire akan yi musu stating dinshi a Turance cikin sauki da fahimtarsu kamar haka:

    Charle’s law states that volume of a given mass of gas at a constant pressure is directly proportional to its absolute temperature. Wannan law din ana amfani da ita ne akan ideal gas held at a constant pressure ta inda volume da temperature ne kawai suke iya canzawa. Gashi kuma a fomula kamar haka:

    V α T or V / T = constant

    Don haka zai zama V1 / T1 = V2 / T

    V1 = initial volume

    T1 = initial absolute temperature

    V2 = final volume

    T2 = final absolute temperature

    Wanda V1 da T1 initial volume ne da initial temperature wato volume da temperature na farko ko muce na ainihi, wanda kuma V2 da Tsune final volume da final temperature na gas gaba daya.

    Akwai abinda ya kamata mu sani wajan lissafin charles law shine temperatures are absolute temperatures, ana measuring dinsu da Kelvin ne ba da  0C ko 0F kuma yana da kyau koda yaushe a rika converting 0C zuwa K ta hanyar adding 273

    Yanzu zamu dauki misalai da zamu amsa su tare da amfani da Charle’s law wanda zamu dauko sune daga jarabawar karshen gama sakandire wato (WAEC,NECO JAMB).

    Za a iya dubaBayanai game da boyle's law

    Misali na daya:

    • The volume of a given mass of gas is 40cm3 at 27 0 what is its volume at 90 0C if its pressure remains constant?      NECO 2003

    Amsar tambaya:

    Zamu fara fitar da data tukunnna

    Data:

    Initial gas volume, V1 = 40cm3

    Initial gas temperature, T1 = 27 0C = (27 + 273) = 300

    Final gas temperature, T= 90 0C = (90 + 273) = 363K

    From V1 / T1 = V2 / T2, Final Volume, V2 = V1T2 / T1 = 40 × 363 / 300 = 48.40cm3

    Ina kara tunasar da ku akan canza 0C zuwa K tare da kara 273 kamar yadda kuka ga na yi wajan amsa wannan tambayar.

    Misali na biyu:

    • Dry oxygen is trapped by a pellet of mercury in a uniform capillary tube which is sealed at one end. The length of the column of oxygen at 27 0C is 50cm. If the pressure of the oxygen is constant, at what temperature will the length be 60cm.        WAEC 1998

    Amsar tambaya: 

    The volume of the gas is assumed to be proportional to the length of the column of gas, i.e V α l

    Initial gas volume, V1 = 50cm

    Initial temperature, T1 = 27 0C = (27 + 273) = 300K

    Final gas volume, V2 = 60cm

    Tunda munfitar da data bari mudauko fomula musa muyi lissafin.

    Daga wannan fomula V1 / T1 = V2 / T2,

                   Final gas volume, T2   = T1V2 /V2 = 300 × 60 / 50 = 360K or 87 0C

    Misali na uku:

    • A fixed mass of gas of volume 600cmat a temperature of 27 0C is cooled at constant pressure to a temperature of 00C. What is the change in volume?              WAEC 1991

    Amsar tambaya:

     Data:

    Initial gas volume, V1 = 600cm3

    Initial gas temperature, T1 = 27 0C = (27 + 273) = 300K

    Final temperature, T2 = 0 0C = (0 + 273) = 273K

    Daga V1 / T1 = V2 / T2  ,

    Final Gas volume, V2 = V1T2 / T1 = 600 × 273 / 300

    = 546cm

    Change in volume = volume at 27 0C  - volume at 0 0

    i.e = V1 – V2 = 600 – 546 = 54cm3

    Duba: Lissafin inclined plane

    Misali na hudu:

    • A gas occupies a certain volume at 27 0C. At what temperature will its volume be doubled assuming that its pressure remains constant? NECO 2000

    Amsar tambaya:

    Data:

    Initial gas temperature, T1 = 27 0C = (27 + 273) = 300K

    Initial gas volume, V1 = V

    Final gas volume, V2 = 2V (Its volume be doubled)

    Zamu sa values din da aka bamu a cikin wannan fomula V1 / T1 = V2 / T2   shine zamu samu

     V / 300 = 2V / T2 therefore, T2 = 300 × 2V / V = 300 × 2 = 600K or 3270C

    Misali na biyar:

    • A gas occupies 221cm3 at a temperature of 0 0C and pressure of 760mm Hg. What will its volume be at 1000C?.

    Amsar tambaya:

    Tunda pressure is constant kuma mass din baya canzawa,don haka munsan zamu iya applying Charle’s law. Idan muka duba tambayar temperature anbamu ne a celcius, don haka yanzu sai mun fara canza su zuwa absolute temperature (Kelvin)

    V1 = 221cm3; T= 273K (0 + 273); T2 = 373K (100 + 273)

    Yanzu zamu iya sa values din a cikin fomula domin samun amsa

    V1 / T1 = V2 / T2

    221cm/ 273K = V2 /373K

    Rearranging the equation to solve for final volume

    V2 = (221cm3) (373K) / 273K

    V2 = 302 cm3

    Wadanan sune kadan daga cikin misalai na lissafin charle’s law wanda yana da kyau mutum ya kara bincike ta hanyar karanta wasu text book na physics sannan kuma tare da amsa wasu tambayoyin domin fahimta da kwarewa akan lissafi.

    Za a iya dubaYadda ake lissafin screw 

    Hakkin mallakar hoto (photo credit): Socratic.org

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