Chủ đề 3: Mô hình hạt của chất
AQA · GCSE · Physics Chủ đề 3 17:01 Giọng đọc tiếng Anh · phụ đề tiếng Anh + 中文 được ghi trực tiếp
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Use mass conservation and density to connect the visible change to a particle explanation.
用质量守恒和密度，把可见变化与粒子解释联系起来。
Do not enlarge the particles themselves.
不要把粒子本身画大。
Ice is a useful exception to the usual solid-liquid density trend.
冰的密度说明固液密度趋势有例外。
Equation-sheet support is determined by the examination series, so practise choosing and rearranging equations.
公式表由考试系列决定，应练习选择和变形公式。
The solid has an ordered close-packed pattern; the liquid is disordered but particles remain close; the gas has much larger separations.
固体排列有序且紧密，液体无序但粒子仍靠近，气体间距大得多。
Solid particles vibrate rather than remaining motionless.
固体粒子在振动，并非静止。
Describe arrangement and motion separately.
分别描述排列和运动。
A gas is less dense mainly because of spacing, not because its particles have become smaller or lighter.
气体密度较小主要是因为间距大，不是粒子变小或变轻。
A ring has a mass of nine point four six grams and a volume of nought point four four cubic centimetres.
一枚戒指的质量为九点四六克，体积为零点四四立方厘米。
Find its density in kilograms per cubic metre.
求它的密度，单位为千克每立方米。
Convert both measurements to SI units first, then state the equation and show the substitution.
先把两个量都换算成国际单位，再写出公式并代入。
Pause and try it.
暂停并试着计算。
A cubic centimetre is a millionth of a cubic metre. Do not use the linear conversion factor for a volume.
一立方厘米是一立方米的百万分之一，体积换算不能用长度换算因子。
The given mass already refers to one second.
已知质量已经对应一秒。
Find the associated volume without introducing a second time factor.
求对应体积，不要再乘或除一次时间。
Keep the requested physical quantity distinct from an extra flow-rate calculation.
把题目要求的体积与额外的流率计算区分开。
Both given units are consistent.
给出的单位已经一致。
For a regular cuboid, measure three actual perpendicular dimensions.
规则长方体要测三个真实的互相垂直尺寸。
For an irregular solid, collect the water displaced by full submersion; use a measuring cylinder to read that volume.
不规则固体完全浸没，收集排出的水，再用量筒读体积。
Plan a method that another student could follow.
写出别人能按顺序完成的方法。
Listing a balance and cylinder without their measurements is incomplete.
只列天平和量筒而不说明测什么，并不完整。
Either displacement method earns credit when the key steps are logically connected.
两种排水法都可以得分，但关键步骤要连贯。
Six marks are judged on the quality of the whole response, not one guaranteed mark per sentence.
六分题按整体方法质量评定，并非每句话保证一分。
Keep the original liquid task separate from the real uncertainty question.
把教师编写的液体题与真题的不确定度题分开。
Subtract the cylinder’s mass before using density.
算密度前，先减去量筒质量。
Give the uncertainty as an interval, and explain what repeat measurements can do.
用区间表示不确定度，并说明重复测量的作用。
A mean reduces random variation; it is not a cure for every error.
平均值能减小随机波动，但不能消除所有误差。
A constant temperature does not mean constant internal energy.
温度不变不等于内能不变。
Use the definition that includes kinetic and potential energy.
要使用同时包含动能和势能的定义。
These linked points match the actual nitrogen explanation.
这些相连步骤对应氮气真题。
For a pure substance at constant pressure, a state-change plateau can occur while energy still transfers.
纯物质在恒压下相变时，即使温度保持不变，仍有能量转移。
Track the vapour that leaves.
追踪离开的水蒸气。
A balance under an open beaker does not measure all material that was initially present.
敞口烧杯下面的天平，并没有测量所有最初存在的物质。
The same conservation reasoning applies to melting, freezing and sublimation when the whole closed system is considered.
把整个封闭系统考虑在内，熔化、凝固和升华也遵守同样的质量守恒。
Choose the equation for a temperature change, not a change of state.
选择温度变化的公式，而非相变公式。
Use final minus initial temperature.
用末温减去初温。
The calculated energy belongs to the pad.
计算得到的是垫子获得的能量。
If an electrical heater also warms the surroundings, its input energy can be larger.
如果电热器还加热了周围环境，输入能量可能更大。
A heating curve for a pure substance, not necessarily water, heated at constant power.
一种纯净物（不一定是水）在恒定功率下加热的升温曲线。
In A the solid warms, in B it melts, in C the liquid warms, in D it boils, and in E the gas warms.
A段固体升温，B段熔化，C段液体升温，D段沸腾，E段气体升温。
During melting and boiling, the state changes with no rise in temperature.
在熔化和沸腾时，物态发生变化而温度不升高。
A plateau describes a process, not a lack of energy input.
平台表示一种过程，不表示没有输入能量。
Compare times only when the energy-transfer rate and latent heat are controlled.
比较时间前，要控制能量转移速率和比潜热。
The boiling plateau is longer in this example because the same mass receives energy at the same rate and its vaporisation latent heat is larger. This is not a universal shape for every material.
在本例中，同样质量以同样速率获得能量，汽化比潜热较大，所以沸腾平台较长；并非所有物质都必须有相同图形。
The mass is given in grams, but latent heat is requested in joules per kilogram.
质量用克给出，比潜热要求焦每千克。
Convert before substituting.
代入前先换算质量。
Nine thousand nine hundred and fifty joules change the state of fifty grams, nought point nought five zero kilograms, of nitrogen.
九千九百五十焦的能量使五十克，即零点零五零千克的氮改变物态。
Use E equals m L, so L equals E over m.
用E等于mL，所以L等于E除以m。
Nine thousand nine hundred and fifty divided by nought point nought five zero is one hundred and ninety-nine thousand joules per kilogram.
九千九百五十除以零点零五零，得每千克十九万九千焦。
Name the correct process, vaporisation for liquid to gas, rather than using a remembered value for water.
要说出正确的过程，液态变气态是汽化，而不要套用记住的水的数值。
The beaker’s mass is present in both readings and cancels.
烧杯质量包含在两次读数中，会相互抵消。
The required mass is the water that evaporated, not either total reading.
公式所需质量是汽化的水，而不是某次总读数。
The mass changed to vapour is nought point nought eight zero minus nought point nought seven one kilograms, which is nought point nought nought nine kilograms; do not use the starting mass of water and beaker.
变成水蒸气的质量为零点零八零千克减去零点零七一千克，即零点零零九千克；不要用水和烧杯的初始质量。
Then L equals E over m: twenty-five thousand two hundred joules divided by nought point nought nought nine kilograms is two point eight million joules per kilogram.
再用L等于E除以m：二万五千二百焦除以零点零零九千克，得每千克二百八十万焦。
Trace whether E or m is too large for the intended boiling process before predicting the direction of error.
先判断针对沸腾过程，E或m哪一个偏大，再判断误差方向。
Each real exam explanation needs the biased quantity and its consequence.
每个真题解释都要说明哪一个量偏差，以及结果方向。
“Heat loss makes it wrong” does not explain the direction.
“散热导致错误”没有解释偏大还是偏小。
The incident molecule can approach at an angle, but the pressure force on the wall is normal to the wall.
分子可以斜着撞向壁面，但压力作用力垂直壁面。
The second panel compares a fixed mass at constant temperature.
第二幅图比较的是恒温下固定质量的气体。
Keep the volume fixed.
保持体积不变。
Temperature is related to average kinetic energy; then connect molecular motion to wall force.
温度与平均动能有关，再把分子运动与壁面受力联系起来。
The mean kinetic energy increases, so the molecules move faster on average.
分子的平均动能增大，分子平均运动得更快。
Each collision exerts a greater force on the walls, and the collisions happen more often.
每次碰撞对器壁的作用力更大，碰撞也更频繁。
So the force per unit area rises, and the pressure increases at fixed volume.
所以单位面积上的力增大，在体积不变时压强增大。
The molecules were already in constant random motion before the gas was heated.
在气体被加热之前，分子就已经在不停地做无规则运动。
Fixed mass and constant temperature are essential.
固定质量和恒定温度是关键条件。
Do not silently use this law for a rapid compression that warms the gas.
不能把这条定律直接用于会升温的快速压缩。
Consistency matters more than converting every number mechanically.
关键是单位一致，不是机械地换算所有数值。
Both pressures use the same unit and both volumes use the same unit.
两端压力单位一致，两端体积单位一致。
Contrast the two experiments by stating what is held constant.
先说明保持不变的条件，再比较两个实验。
“Pressure rises because it gets hotter” is not the explanation for isothermal compression.
恒温压缩时，不能说压力因为温度升高而增加。
The moving piston transfers energy by work.
运动的活塞通过做功转移能量。
This case has different conditions from the preceding constant-temperature syringe calculation.
这里的条件不同于上一道恒温注射器计算。
Keep work and heating as distinct energy-transfer routes.
把做功和热传递区分为两种能量转移方式。
The syllabus asks why work on a gas can warm it, not a universal claim that compression always heats.
考纲要求解释做功为什么可能使气体升温，不是说所有压缩都必定升温。
Use the questions to locate the sheet you need to revisit.
用这些问题找到需要复习的练习。
Explain the conditions and the measured quantities before reaching for an equation.
先解释条件和测量量，再选择公式。
A correct equation with the wrong process or condition is not a complete solution.
公式正确，但过程或条件错误，不能构成完整解答。
Revisit any answer you could recognise but could not explain.
对那些能认出却无法解释的答案，再做一次。