Thermodynamique et mécanique statistique
GRE Subject Test · GRE Physics Sujet 4 3:55 Narration en anglais · Sous-titres anglais + 中文 incrustés
Chapitres
Transcript
Two engines can both conserve energy, yet only one respects the entropy limit on efficiency.
两台热机都可以遵守能量守恒，但只有一台满足熵对效率的限制。
This topic applies the first and second laws with a clear sign convention, relates ensembles, Boltzmann factors and entropy, and analyses ideal gases, heat engines and quantum statistics.
本主题按明确的符号约定应用热力学第一、第二定律，联系系综、玻尔兹曼因子和熵，并分析理想气体、热机和量子统计。
State the first-law sign convention: ΔU=Q−W when W is work done by the system.
陈述热力学第一定律的符号约定：ΔU = Q − W，其中 W 为系统对外做的功。
For a reversible engine between reservoirs, efficiency is bounded by 1−Tc/Th using absolute temperatures.
对于在两个热库之间工作的可逆发动机，其效率受限于 1 减去 T 下标 c 除以 T 下标 h，此处使用绝对温度。
Boltzmann weights are exp(−E/kT).
玻尔兹曼权重为 e 的负 E 除以 kT 次方。
Probabilities require division by the partition function.
概率计算需要除以配分函数。
Distinguish Maxwell–Boltzmann, Bose–Einstein and Fermi–Dirac assumptions.
区分麦克斯韦-玻尔兹曼、玻色-爱因斯坦和费米-狄拉克的假设。
Fermions cannot share an identical single-particle state.
费米子不能占据相同的单粒子态。
Take the work W done by the gas as positive.
规定气体对外做的功 W 为正。
Then the first law says the change in internal energy, delta U, equals the heat Q put in, minus the work W the gas does.
于是热力学第一定律表示：内能的变化 ΔU 等于吸收的热量 Q 减去气体对外做的功 W。
For a reversible isothermal expansion of an ideal gas, U depends only on T, so delta U is zero and the heat in equals the work out.
对理想气体的可逆等温膨胀，内能 U 只取决于温度 T，所以 ΔU 为零，吸收的热量等于对外做的功。
The work is the integral of p d V, which is n R T times the log of V final over V initial.
功等于 p dV 的积分，即 nRT 乘以末体积与初体积之比的自然对数。
For half a mole at 300 kelvin, doubling the volume: W equals n R T ln 2, which is 0.50 times 8.314 times 300 times ln 2, about 864 joules.
取 0.50 摩尔气体，温度 300 K，体积加倍：W 等于 nRT ln2，即 0.50 乘 8.314 乘 300 再乘 ln2，约为 864 焦耳。
The gas's entropy change is the reversible heat over the temperature: 864 joules over 300 kelvin, which is 2.88 joules per kelvin.
气体的熵变等于可逆过程吸收的热量除以温度：864 焦耳除以 300 K，等于每开尔文 2.88 焦耳。
The gas expands and receives heat.
气体膨胀并吸收热量。
The reservoir loses the same entropy in this reversible limit.
在此可逆极限下，热源失去相同的熵。
An insulated rigid vessel contains ideal gas confined initially to half its volume.
一个绝热刚性容器内装有理想气体，初始时被限制在容器体积的一半中。
Remove the internal partition into vacuum.
移除内部隔板使其向真空自由膨胀。
Compare Q, W, temperature change and gas entropy change with reversible isothermal expansion to the same final volume.
请将其与可逆等温膨胀至相同终态体积的过程进行比较，分析热量Q、功W、温度变化以及气体熵变。
Free expansion has Q equals zero by insulation and W equals zero because external pressure is zero.
自由膨胀中，Q等于零，因为绝热；W等于零，因为外压为零。
Thus, delta U equals zero and ideal-gas temperature is unchanged.
因此，ΔU等于零，理想气体温度不变。
Entropy is a state function: delta S equals n R ln 2, which is greater than zero, evaluated on an imagined reversible isothermal path.
熵是状态函数：ΔS等于nRln2，大于零，该值在假想的可逆等温路径上计算得出。
The actual process is irreversible; Q sub actual divided by T equals zero is not its entropy change.
实际过程是不可逆的，Q_实际除以T等于零，并非其熵变。
Reversible expansion has Q equals W equals n R T ln 2 and the same gas entropy increase, with a matching reservoir entropy decrease.
可逆膨胀中，Q等于W等于nRTln2，气体熵增相同，同时伴有储热库熵减相匹配。
Award the four quantities and comparison.
请给出这四个量及其比较结果。
Celsius temperature ratios cannot be used in the Carnot efficiency formula.
摄氏度温度比值不能用于卡诺效率公式。
entropy: A state quantity connected to multiplicity and reversible heat transfer.
熵：与多重性和可逆热传递相关的状态量。
partition function: The sum of statistical weights used to normalise probabilities.
配分函数：用于归一化概率的统计权重之和。
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其含义，然后改变原始传递中的一个参数并检验极限情况。