onWindow":[{"id":"25aaea9d-bba9-4769-9ba0-a4d82087a659","revisionId":"b456d6ac-1c4e-4ab0-b96e-ee47b9f0e1f6","questionNumber":null,"questionType":"LA","specification":"A cylinder contains a fixed mass of gas sealed by a moveable piston.","difficulty":5,"difficultyBand":"medium","totalMarks":5,"sourceTitle":null,"calculatorUse":null,"parts":[{"id":"3e08adb4-222f-48e3-8d5d-8ab76ce2365a","label":"a","content":"The volume of the gas is initially \$1.8 \\times 10^{-4}\\text{ m}^3\$ and its pressure is \$2.0 \\times 10^4\\text{ Pa}\$. The piston is slowly pushed inside the cylinder, keeping the temperature of the gas constant, until the volume of the gas is reduced to \$4.5 \\times 10^{-5}\\text{ m}^3\$.\n\nCalculate the new pressure of the gas.","markscheme":"- Uses $P_1 V_1 = P_2 V_2$ to calculate the constant: $2.0 \\times 10^4 \\times 1.8 \\times 10^{-4} = 3.6\\text{ Pa m}^3$ M1\n- Calculates the new pressure: $P_2 = \\frac{3.6}{4.5 \\times 10^{-5}} = 8.0 \\times 10^4\\text{ Pa}$ A1","marks":2,"order":0,"markingRubric":null,"explanation":{"methods":[{"id":"boyles_law","label":"Use Boyle's law","steps":[{"type":"text","order":0,"title":"Choose the relationship","content":"Because the gas has a fixed mass and the temperature stays constant, pressure multiplied by volume stays constant. Use Boyle’s law:\n\n$$P_1V_1=P_2V_2$$\n\nHere, subscript 1 means **initial** and subscript 2 means **final**.","highlights":[{"text":"fixed mass of gas","location":"specification"},{"text":"keeping the temperature of the gas constant","location":"specification"}]},{"type":"hyper_latex","block":{"nodes":[{"latex":"\\htmlData{anchor=initial}{\\color{@sky}{P_1V_1}}=P_2V_2"},{"latex":"\\htmlData{anchor=substitution}{\\color{@sky}{(2.0\\times10^{4}\\,\\mathrm{Pa})(1.8\\times10^{-4}\\,\\mathrm{m^3})}}=P_2V_2"},{"latex":"\\htmlData{anchor=constant}{\\color{@amber}{3.6\\,\\mathrm{Pa\\,m^3}}}=P_2V_2"}],"highlights":[{"color":"sky","style":"text-color-with-background","anchor":"initial"},{"color":"amber","style":"text-color-with-background","anchor":"constant"}]},"order":1,"title":"Find the constant","description":"M1: using P1V1 = P2V2 and calculating the constant gives 3.6 Pa m³."},{"type":"hyper_latex","block":{"nodes":[{"latex":"\\htmlData{anchor=known}{\\color{@sky}{3.6\\,\\mathrm{Pa\\,m^3}}}=P_2(4.5\\times10^{-5}\\,\\mathrm{m^3})"},{"latex":"\\htmlData{anchor=divide}{\\color{@sky}{P_2}}=\\dfrac{3.6}{4.5\\times10^{-5}}\\,\\mathrm{Pa}"},{"latex":"\\htmlData{anchor=answer}{\\color{@amber}{P_2=8.0\\times10^{4}\\,\\mathrm{Pa}}}"}],"highlights":[{"color":"amber","style":"text-color-with-background","anchor":"answer"}]},"order":2,"title":"Calculate the new pressure","description":"A1: calculating the final pressure gives 8.0 × 10⁴ Pa."}]}],"generatedAt":"2026-07-22T02:17:12.653Z","modelVersion":"gpt-5.6-terra+diagrams-1","explanationVersion":"v3"},"options":[]},{"id":"f21d31c9-a9e5-4694-ad0a-db031d7a5e57","label":"b","content":"Explain, in terms of the behavior of gas particles, why the pressure inside the cylinder increases as the volume is decreased.","markscheme":"- States that the average speed (or kinetic energy) of the gas particles remains constant because the temperature is constant B1\n- Explains that decreasing the volume reduces the distance particles travel between the walls, leading to more frequent collisions (a higher rate of collisions) with the container walls B1\n- Explains that more frequent collisions result in a greater total force per unit area on the walls, which increases the pressure B1","marks":3,"order":1,"markingRubric":null,"explanation":{"methods":[{"id":"particle_collision_explanation","label":"Explain particle collisions","steps":[{"type":"text","order":1,"title":"Keep speed constant","content":"Assuming the temperature remains constant, the particles’ average speed (and therefore their average kinetic energy) stays constant. This earns the first B1 mark."},{"type":"text","order":2,"title":"Explain collision frequency","content":"When t

Pressure

Question 6

Pressure Questions

Pressure

Question 6

Pressure Questions