Gas processes, entropy and reversible cycles
GRE Subject Test · GRE Physics Topic 19 6:14 English narration · English + 中文 subtitles burned in
Chapters
Transcript
A refrigerator can move more heat than the work supplied to it.
冰箱搬运的热量可以多于输入给它的功。
That ratio is a coefficient of performance, not an efficiency that must stay below one.
这个比值是性能系数，而不是必须小于一的效率。
This topic applies first-law signs to gas processes, compares reversible entropy balances and P V cycle areas, and separates engine efficiency from a refrigerator's coefficient of performance.
本主题对气体过程应用热力学第一定律的符号，比较可逆过程的熵平衡和 P-V 循环面积，并区分热机效率与冰箱的性能系数。
Use Delta U equals Q minus W, where Q is heat entering the system and W is work done by it.
使用 ΔU = Q − W，其中 Q 为进入系统的热量，W 为系统对外做的功。
Quasistatic boundary work is the integral of P dV, positive during expansion.
准静态边界功为 ∫P dV，膨胀时为正。
For a fixed amount of ideal gas, internal energy depends only on temperature.
对于一定质量的理想气体，内能仅取决于温度。
At constant volume, W equals zero, so Delta U equals Q. For an isothermal ideal-gas process, Delta U equals zero, so Q equals W.
等容过程中 W = 0，故 ΔU = Q；等温理想气体过程中 ΔU = 0，故 Q = W。
Adiabatic means Q equals zero, not constant temperature.
绝热指 Q = 0，并非温度恒定。
Over a cycle, the state returns and Delta U equals zero, so net Q equals net W even though individual legs have different heat transfers.
经历一个循环后状态复原，ΔU = 0，因此净热量等于净功，尽管各分过程的热量传递不同。
For reversible isothermal expansion of an ideal gas, P V is constant.
对于理想气体的可逆等温膨胀，P V 为常数。
With constant heat capacities, a reversible adiabatic path obeys P V to the gamma equals constant, with gamma equals C sub p divided by C sub v greater than one.
在热容恒定的情况下，可逆绝热路径满足 P V 的 γ 次方等于常数，其中 γ 等于 C_p 除以 C_v，且大于 1。
Through the same initial state, its pressure falls faster as volume increases.
经过相同的初始状态时，随着体积增加，其压力下降得更快。
An isobaric path is horizontal on a P dash V plot.
等压路径在 P-V 图上呈水平线。
A clockwise closed cycle has positive integral of P dV around the loop; reversing direction changes the sign.
顺时针闭合循环具有正的 ∮P dV；反转方向会改变符号。
For a rectangular loop, the magnitude is delta P times delta V, and a triangular loop has half the corresponding bounding rectangle area.
对于矩形回路，其大小为 ΔP 乘以 ΔV；三角形回路的面积是对应外接矩形面积的一半。
Use Pa and m cubed for joules, not an unconverted litre value.
使用 Pa 和 m³ 来计算焦耳，不要使用未换算的升值。
For reversible heat transfer, d S equals delta Q sub rev divided by T.
对于可逆热传递，dS等于δQ_rev除以T。
Entropy is a state function while heat is path dependent.
熵是状态函数，而热量取决于路径。
A reversible system plus environment process has zero total entropy change, but the system's entropy alone can increase or decrease.
可逆的系统加环境过程的总熵变为零，但系统自身的熵可以增加或减少。
Irreversible spontaneous processes produce nonnegative total entropy.
不可逆自发过程产生非负的总熵。
A reversible isothermal expansion increases the gas entropy by n R ln of V2 over V1; the reservoir loses the same amount.
可逆等温膨胀使气体熵增加nR乘以ln(V2/V1)；储热源损失相同的量。
Reversibility does not require constant system temperature, zero work or zero internal energy change.
可逆性不要求系统温度恒定、做功为零或内能变化为零。
A reversible engine between T hot and T cold has efficiency W over Q hot, equal to 1 minus T cold over T hot.
在 T 高与 T 低之间工作的可逆发动机，效率为 W 除以 Q 高，等于 1 减去 T 低除以 T 高。
A reversible refrigerator instead has a coefficient of performance Q cold over W, equal to T cold over the difference T hot minus T cold, and a heat pump has Q hot over W, equal to T hot over that same difference.
可逆制冷机的性能系数为 Q 低除以 W，等于 T 低除以 T 高与 T 低之差；热泵的 Q 高除以 W 等于 T 高除以同一个差。
Use absolute kelvin temperatures.
请使用绝对开尔文温度。
On a reversible T–S diagram, heat magnitude along an isotherm is T times the entropy change; the Carnot rectangle's area is the work magnitude.
在可逆 T-S 图上，等温线上的热量大小等于 T 乘以熵变；卡诺矩形的面积即为功的大小。
Refrigeration moves heat from cold to hot by consuming work, reversing the engine cycle.
制冷过程通过消耗功将热量从低温移向高温，从而反转发动机循环。
The coefficient of performance can exceed one without violating conservation because the moved heat is not supplied solely by the work.
性能系数可以超过 1 而不违反能量守恒定律，因为被转移的热量并非仅由功提供。
A reversible refrigerator moves 3.0 joules per kelvin of entropy.
一台可逆冰箱搬运每开尔文 3.0 焦耳的熵。
From the cold side at 270 kelvin it takes 270 times 3.0: 810 joules.
它从 270 开尔文的冷端吸收 270 乘 3.0：810 焦耳。
It rejects the same entropy at 300 kelvin: 300 times 3.0, which is 900 joules.
它在 300 开尔文处放出同样的熵：300 乘 3.0，即 900 焦耳。
The work input is the difference: 900 minus 810, which is 90 joules.
输入的功是两者之差：900 减 810，等于 90 焦耳。
The coefficient of performance is 810 over 90, which is 9.
性能系数等于 810 除以 90，即 9。
A value above one is allowed because the heat is moved, not produced from the work.
它可以大于一，因为热量是被搬运的，而不是由功产生的。
A refrigerator removes 600 J from 280 K and rejects 650 J to 300 K per cycle.
一台冰箱从 280 K 的热源移除 600 J 的热量，并向 300 K 的热源排出 650 J 的热量，每循环一次。
Check energy balance and total entropy change.
请检查能量平衡和总熵变。
Is the proposed cycle possible?
该提议的循环是否可能实现？
Find the minimum work needed for the same cooling between these reservoirs.
求在相同冷源与热源之间进行同等冷却所需的最小功。
Energy requires 50 J work input.
能量需要 50 J 的功输入。
Reservoir entropy change is 650 divided by 300 minus 600 divided by 280, which equals plus 0.02381 joules per kelvin, which is greater than zero, so the complete cycle does not violate the second law.
热库熵变为 650 除以 300 减去 600 除以 280，等于正 0.02381 J/K，大于零，因此完整循环不违反第二定律。
Reversible minimum work is Q sub c times T sub h divided by T sub c minus 1, which equals 600 times 300 divided by 280 minus 1, which equals 42.86 joules.
可逆最小功为 Q_c(T_h/T_c - 1)，即 600 乘以 (300/280 - 1)，等于 42.86 J。
The proposed cycle is possible thermodynamically and uses more than minimum work.
该提议循环在热力学上是可能的，且使用的功多于最小功。
Award work, correct entropy arithmetic, sign possibility, reversible bound and comparison.
评分要点：功、正确的熵计算、符号与可能性、可逆界及比较。
Zero total entropy production does not mean zero system entropy change.
总熵产为零并不意味着系统熵变为零。
A refrigerator uses TC/(TH−TC), not the engine efficiency formula.
制冷机使用T下C除以T下H减T下C，而非热机效率公式。
entropy production: Nonnegative total entropy generated by irreversibility.
熵产生：由不可逆性产生的非负总熵。
coefficient of performance: Useful heat transferred divided by work input for a refrigerator or heat pump.
性能系数：用于制冷机或热泵的有用传热量除以输入功。
Retrieve the meanings without reading the definitions, then change a parameter in the original transfer and test the limiting case.
在不阅读定义的情况下回忆含义，然后改变原始传递过程中的一个参数，并检验极限情况。