Website content © Dr Phil Brown 2000+. All copyrights reserved on Doc Brown's Chemistry revision notes, images, quizzes, worksheets etc. Copying of website material is NOT permitted. GCSE level and advanced pre-university level revision notes. Detailed notes on the states of matter and their properties. Based on the syllabus-specifications for students taking the IGCSE/GCSE level physics examinations summary revision notes and key points about more complex P-V-T calculations, for students taking the WJEC gcse chemistry/physics, CCEA gcse chemistry/physics, CIE igcse chemistry/physics, AQA igcse/gcse physics, Edexcel gcse chemistry/physics, OCR 21st century chemistry/physics, OCR gateway chemistry/physics or any other GCSE or IGCSE level chemistry/physics exams e.g. US grade 9-10 physics courses INDEX of all my notes on the states of matter GCSE (~US grades 8-10) level multiple choice QUIZ on the states of matter: gases, liquids & solids

ANSWERS to the more complex P-V-T calculations

• Q1

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

  • 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 25^oC before combustion and rise to a peak temperature of 2100^oC 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 20^oC 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 = 8^oC

• 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 100^oC and a pressure of 150 kPa.

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

  • 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 22^oC 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 27^oC (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 27^oC,  would occupy at 22^oC.

    • 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 22^oC be used in the rate calculation analysis? and are this still other sources of error?

    • The theoretical–calculated gas volume for 22^oC 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. 27^oC) 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 27^oC, that would have occurred/been measured at 22^oC 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 27^oC and there isn't much you can do about that error except repeat the experiment at 22^oC, 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.