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STATES OF MATTER - properties of gases and liquids (fluids) and solids

20. The combined gas law equation - more complicated P-V-T calculations P1V1T1 = P2V2T2

Doc Brown's chemistry revision notes: basic school chemistry science GCSE chemistry, IGCSE  chemistry, O level and ~US grades 8, 9 and 10 school science courses or equivalent for ~14-16 year old science students for national examinations in chemistry and also helpful for UK advanced level chemistry students aged ~16-18 and US grades 11-12 K12 honors.


20. The combined gas law equation - more complicated P-V-T calculations

  • If all the laws described in 4b and 4c are combined, you get the following general expression

  • p x V/T = a constant (for a given mass of gas).

  • This can be expressed in generalised form for calculations based on an initial set of conditions1 (1) changing to a new and final set of conditions2 (2) for a given mass of gas, giving the combined pressure没olume釦emperature gas calculation equation ...

  • p1 x V1   p2 x V2
    末末末末末末 = 末末末末末末
    T1   T2
  • In shorthand': p1V1 /T 1 = p2V2 / T2
  • therefore the three permutations for problem solving involving all three variables are:
    • p2 = p1V1T2 / V2T1
    • V2 = p1V1T2 / p2T1
    • T2 = p2V2T1 / p1V1
  • If one variable is constant, the permutations for the other two variables are:

    • p2 = p1V1 / V2   (at constant temperature)
    • V2 = V1T2 / T1   (at constant pressure)
    • T2 = p2T1 / p1   (at constant volume)
  • Note:

    • If the temperature is constant you get Boyle's Law.

    • If p or V is constant you get Charles's/Gay豊ussac's Law.

    • You can use any volume or pressure units you like as long as both p's or both V's have the same units.

    • The graphs of p or V versus temperature become invalid once the gas has condensed into a liquid BUT when extrapolated back all the lines seem to originate from y = 0 (for p or V), x = 273oC (for T).

    • This was part of the scientific evidence that led to the belief that 273oC was the lowest possible temperature, though there is no theoretical upper limit at all.

    • This led to the devising of a new thermodynamic absolute temperature scale or Kelvin scale which starts at OK.

      • e.g. ice melts at 0oC or 273K and water boils at 100oC or 373K.

  • Examples of P坊傍 calculations for you to practice (ANSWERS at end of page)

  • Q1

    • 25 cm3 of a gas at 1.01 atm. at 25oC was compressed to 15 cm3 at 35oC.

    • Calculate the final pressure of the gas.

    • -

  • Q2

    • The fuel and air gases in the cylinders of a 1200 cm3 car engine go from 25oC before combustion and rise to a peak temperature of 2100oC after combustion. If normal atmospheric pressure is 101 kPa, calculate the peak pressure reached after combustion. Although the movement of the piston changes the volume, for the sake of argument (i) assume the volume is constant at both ends of the cycle.

    • -

    • (ii) To be more realistic, assume the initial volume of fuel vapour plus air was 400 cm3, now re-calculate the final pressure.

    • You now need to use the full PVT expression.

    • -
  • Q3

    • A 5 dm3 (5 litre) cylinder of a gas was stored at 20oC with an internal gas pressure of 100 kPa.

    • At what temperature in oC would the pressure be 120 kPa if the volume of the storage cylinder was reduced to 4 dm3 (4 litre).

    • -

  • Q4

    • Prior to compression, the internal volume of one cylinder of a car engine is 250 cm3.

    • Assume the air-fuel mixture is at a temperature of 100oC and a pressure of 150 kPa.

    • After piston compression and ignition the gas volume is reduced to 50 cm3 and the temperature rises to 800oC.

    • Calculate the pressure created after compression and ignition.

    • -
       

  • Q5

    • A student was investigating the speed of reaction between limestone granules and different concentrations of hydrochloric acid. However after doing a whole series of experiments at different acid concentrations, there was no time to do the last planned experiment. The volume of carbon dioxide collected after 5 minutes in a 100cm3 gas syringe was used to determine the rate of reaction. All the experiments were done in one lesson at a temperature of 22oC except for the last one. This was done in the next lesson, giving a carbon dioxide volume of 47.0 cm3 after 5 minutes, but at a higher temperature of 27oC (when in Kelvin call this T1, and the other temperature T2).

    • To make the data analysis fair, all the gas volumes should be ideally measured at the same temperature, but a correction can be made for the last experiment.

    • (a) Calculate the volume the of 47.0 cm3 of gas at 27oC,  would occupy at 22oC.

      • -

    • (b) If the temperature was ignored, what is the % error in the rate of reaction measurement?

      • -

    • (c) Should the calculated value for 22oC be used in the rate calculation analysis? and are this still other sources of error? Discuss!!!

      • -

    • (d) Would you need to do any correction for the volume of acid added to the limestone? Explain your decision.


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Detailed notes on the states of matter and their properties

ANSWERS to the more complex P-V-T calculations

  • Q1

    • 25 cm3 of a gas at 1.01 atm. at 25oC was compressed to 15 cm3 at 35oC.

    • Calculate the final pressure of the gas.

    • p1 = 1.01 atm, p2 = ?, V1 = 25 cm3, V2 = 15 cm3,

    • T1 = 25 + 273 =  298 K, T2 = 35 + 273 = 308 K

    • (p1 x V1)/T1 = (p2 x V2)/T2

    • p2 = (p1 x V1 x T2)/(V2 x T1)

    • p2 = (1.01 x 25 x 308)/(15 x 298) = 1.74 atm

    • -

  • Q2

    • The fuel and air gases in the cylinders of a 1200 cm3 car engine go from 25oC before combustion and rise to a peak temperature of 2100oC after combustion. If normal atmospheric pressure is 101 kPa, calculate the peak pressure reached after combustion. Although the movement of the piston changes the volume, for the sake of argument (i) assume the volume is constant at both ends of the cycle.

    • T1 = 25 + 273 = 298 K, T2 = 2100 + 273 = 2373 K, P1 = 101 KPa

    • p/T = constant

    • p1/p2 = T1/T2

    • p2 = p1 x T2/T1

    • p2 = 101 x 2373/298 = 804 kPa

    • (ii) To be more realistic, assume the initial volume of fuel vapour plus air was 400 cm3, now re-calculate the final pressure.

    • You now need to use the full PVT expression.

    • (p1 x V1)/T1 = (p2 x V2)/T2
    • p2 = (p1 x V1 x T2)/(V2 x T1)
    • p2 = (101 x 400 x 2373)/(1200 x 298) = 268 kPa
  • Q3

    • A 5 dm3 (5 litre) cylinder of a gas was stored at 20oC with an internal gas pressure of 100 kPa.

    • At what temperature in oC would the pressure be 120 kPa if the volume of the storage cylinder was reduced to 4 dm3 (4 litre).

    • T2 = P2V2T1 / P1V1

    • T2 = (120 x 4 x(20 + 273)) / (100 x 5)

    • T2 = 14060 / 500 = 281.3 K

    • T2 = 281 K (3 sf)

    • T2 = 281 - 273 = 8oC

  • Q4

    • Prior to compression, the internal volume of one cylinder of a car engine is 250 cm3.

    • Assume the air-fuel mixture is at a temperature of 100oC and a pressure of 150 kPa.

    • After piston compression and ignition the gas volume is reduced to 50 cm3 and the temperature rises to 800oC.

    • Calculate the pressure created after compression and ignition.

    • P2 = P1V1T2 / V2T1

    • P2 = (150 x 250 x (800 + 273)) / (50 x (100 + 273))

    • P2 = 40237500 / 18650 = 2157.5

    • P2 = 2160 kPa (3 sf)
       

  • Q5

    • A student was investigating the speed of reaction between limestone granules and different concentrations of hydrochloric acid. However after doing a whole series of experiments at different acid concentrations, there was no time to do the last planned experiment. The volume of carbon dioxide collected after 5 minutes in a 100cm3 gas syringe was used to determine the rate of reaction. All the experiments were done in one lesson at a temperature of 22oC except for the last one. This was done in the next lesson, giving a carbon dioxide volume of 47.0 cm3 after 5 minutes, but at a higher temperature of 27oC (when in Kelvin call this T1, and the other temperature T2).

    • To make the data analysis fair, all the gas volumes should be ideally measured at the same temperature, but a correction can be made for the last experiment.

    • (a) Calculate the volume the of 47.0 cm3 of gas at 27oC,  would occupy at 22oC.

      • V1/V2 = T1/T2 so V2 = V1 x T2/T1

      • V1 = 47.0 cm3, T1 = 273 + 27 = 300K, T2 = 273 + 22 = 295K

      • V2 = 47.0 x 295/300 = 46.2 cm3

    • (b) If the temperature was ignored, what is the % error in the rate of reaction measurement?

      • Volume error = 47.0 46.2 = +0.8 cm3, therefore ....

      • % error = 0.8 x 100/47 = +1.7% (so you would over calculate the reaction rate without this correction)

      • The % error in the volume would be the same as calculated for the rate e.g. in cm3/min.

    • (c) Should the calculated value for 22oC be used in the rate calculation analysis? and are this still other sources of error?

      • The theoretical膨alculated gas volume for 22oC should be used for calculating the rate, it will improve the accuracy a little, BUT there is another problem!

      • If the reaction was unfortunately carried out at a higher temperature (i.e. 27oC) there is a second source of error. At a higher temperature the reaction is faster, so you are bound to get a higher volume of gas formed in five minutes. Therefore you will calculate a faster rate of reaction e.g. in cm3 gas/minute at 27oC, that would have occurred/been measured at 22oC and so an unfair comparison with all the other results from the previous lesson.

      • So, although you can correct reasonably well the volume error due to an 'expanded' gas volume at the higher temperature, the gas volume will still be too high because of the faster rate of reaction at 27oC and there isn't much you can do about that error except repeat the experiment at 22oC, which is the best thing to do anyway!

        • Note that if the temperature of a rates experiment was too low compared to all the other experiments, the 'double error' would occur again, but this time the measured gas volume and the calculated speed/rate of reaction would be lower than expected.

    • (d) Would you need to do any correction for the volume of acid added to the limestone? Explain your decision.

      • No correction needed for this at all. Although liquids expand/contract on heating/cooling, the volume changes are far less compared to gas volume changes for the same temperature change. This is because of the relatively strong intermolecular forces between liquid molecules, which are almost absent in gases.

 

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