Makalu

Pressure Law: Ma'anarsa da yadda ake lissafin shi

  • Pressure law yana 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 ne kai tsaye dan duba daya daga cikin gas laws, wannan law din shine pressure law, ana bayanin law dinne a Turance kamar haka:

    Pressure law states that the pressure of a fixed mass of gas at constant volume is proportional to the absolute temperature of the gas. Idan muka koma bangaran lissafi kuma akanyi bayanin sa ne da wadannan fomula da zan kawo kamar haka:

    P α T or P / T = constant

    Sannan kuma, P1 /T1 = P2/T2

    P= initial gas pressure

    T= initial gas temperature

    P= final gas pressure

    T= final gas temperature.

    Yana da kyau mu sani cewa idan zamu yi lissafi a karkashin pressure law in aka bamu temperature in degree Celsius (0C) mu canza shi zuwa Kelvin (K) idan zamu yi applying pressure law. Yanzu zamu dauki misalai na lissafin pressure law mu amsa su wanda kadan ne daga cikin tambayoyin da ke fitowa a jara bawar karshen gama sakandire wato kamar su WAEC, JAMB, NECO.

    Misali na daya:

    • Before starting a journey, the tyre pressure of a car was 3 × 105Nm-2 at 270 At the end of the journey, the pressure rose 4 × 105Nm-2 . Calculate the temperature of the tyre after the journey assuming the volume is constant.
    1. 4000C, B. 3000C, C. 2730C,  1270C(JAMB 1997)

    Fahimtar abinda tambaya take nufi ko take bukata wajan amsata yana da mahimmanci saboda shi Physics ba kamar Maths ba ne da za a baka tambaya kasancewa ga formula da za kai amfani da ita, dole sai ka gane inda tambayar ta dosa, kamar yadda muka gani an gaya mana cewa kafin a fara tafiya da motar pressure da ke cikin tayar mota ana gaya mana ita  tare da degree dinta sannan kuma aka gaya mana a karshen tafiyar nawa ne pressure sannan aka bukaci mu yi lissafin temperature tayar a karshen tafiyar, idan mu kai duba da fomular mu an bamu dukkan abubuwan banda final temperature yanzu zamu fara fitar da data.

    Data:

    Initial gas pressure P1 = 3 × 105Nm-2

    Initial gas temperature, T1 = 270C = (27 + 273) = 300K

    Final gas pressure, P2 = 4 × 105Nm-2

    Daga P1/ T1 = P / T, final temperature, T2 = P2 T1 / P1

    Yanzu tunda mun fitar da fomula zamu dauko abubuwan da aka bamu musa a ciki.

    T= 4 × 10× 300 / 3 × 10= 400K or (400 – 273 )0C = 1270C

    Misali na biyu:

    • A gas at pressure P Nm-2 and temperature 270C is heated to 770C at constant volume. What is the new pressure? 0.85 PNm-2   B.0.86 PNm-2    C.1.16 PNm-2   D. 1.18PNm-2

    E.2.85P Nm-2 (JAMB 1978)

    Amsar tambaya: 

    Data:

    Initial gas pressure, P1 = PNm-2

    Initial gas temperature, T1 = 270C = (27 + 273) = 300K

    Final gas temperature, T2 = 770C = (77 + 273) = 350K

    From P1 / T1 = P/T2,

    Final pressure, P= P1T2 / T= 1.16 PNm-2

    Dubi:  Physics: Bayanai game da gas laws

    Misali na uku:

    • A closed inexpansible vessel contains air saturated with water vapour at 770C. the total pressure in the vessel is 1007 mmHg. Calculate the new pressure in the vessel if the temperature is reduced to 270 (S.V.P of water at 770C and 270C respectively are 314mmHg anc 27mmHg. Treat the air in the vessel as an ideal gas). WAEC2008/12

    Amsar tambaya: 

    Kamar yadda yazo a Law din Partial Pressure a Turance gashi kamar haka: The pressure of air alone in the vessel is equal to the difference between the total pressure in the vessel and the S.V.P. of water at the same temperature.

    Yanzu zamu fitar da data:

    Data:

    Initial air pressure, P1 = 1007 – 314 = 693mmHg

    Final air pressure, P2 =?

    Initial temperature, T1 = 770C = 77 + 273 = 350K

    Final temperature, T2 = 270C = 27 + 273 = 350K

    Because the vessel is inexpansible, its volume does not change, therefore pressure Law can be applied.

    Substitute into P1 / T1 = P/ Tto obtain

    693/350 = P/ 300

    P= 693 × 300 / 350 = 594mmHg.

    The pressure in the vessel at 270C is equal to the sum of the pressure of air and the S.V.P. of water at 270C.

    Vessel pressure at 270C = air + S.V.P of water at 270C

    = 594mmHg + 27 mmHg

    = 621 mmHg

    Misali na hudu:

    • If the pressure of a gas is 70 cmof mercury at 200C, determine its pressure at 500C, assuming that the volume is constant.

    Amsar tambaya:

    P1 / T1 = P/ T 

    70 / 273 + 20 = P2 / 273 + 50

    P= 70 × 324 / 293

    = 77.17 cm3

    Wadannan sune kadan daga cikin misalan pressure law wanda dalibai yana da kyau su kara duba wasu textbooks din domin karatu da kuma kokari wajan amsa wasu tambayoyi na daban a karkashin wannan topic din domin samun cikkaken fahimta.

    Sannan za ku iya duba makalu na wanda suka shafi gas law, da boyle’s law, da charle’s law da sauransu. Sannan da yardan Allah, makala ta da zanyi gaba ita ce Ideal Gas law wadda aka fi sani da general gas law. Na gode, sai mun hadu a makala ta gaba.

    Hakkin mallakar hoto (photo credit): passmyexams.co.uk

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