The first law for closed systems: work is area
Введено| English | Значение |
|---|---|
| closed system/kləʊzd ˈsɪstəm/ | closed system |
| boundary work | Work transferred by movement of a system boundary; quasistatic boundary work is the integral of pressure with respect to volume. |
| path dependence | path dependence |
| isothermal process | isothermal process |
| adiabatic process | adiabatic process |
| state function/steɪt ˈfʌŋkʃn/ | state function |
The accountant's law
- A closed system 闭口系统 has no mass crossing its boundary, and the first law is its ledger: $\Delta U = Q - W$, with work by the system positive. Energy changes only by heat and work crossing the boundary - nothing else.
- Boundary work 边界功 is $W = \int_1^2 p\,dV$, the area under the path on a p-V diagram. Because the integral runs along the path $p(V)$, work is a path dependence 路径依赖 quantity: same endpoints, different paths, different answers.
Which of these is a state function?
Сложность 1/5 · 100 монет + небольшой бонус за первое завершение сегодня
Work and heat are path quantities; h = u + pv depends only on the state.
Three paths, one pair of states
- Air starts at $(200\ \mathrm{kPa}, 0.3\ \mathrm{m^3})$ and triples its volume. An isobaric process gives $W = p\Delta V = 200 \times 0.6 = \mathbf{120}\ \mathrm{kJ}$. An isothermal process 等温过程 through the same ends gives $W = p_1 V_1 \ln 3 = \mathbf{65.9}\ \mathrm{kJ}$ - the same endpoints, $54.1\ \mathrm{kJ}$ apart, and the difference is exactly the area between the two curves.
- Now an adiabatic process 绝热过程 ($Q = 0$): compress from the same start to $0.15\ \mathrm{m^3}$: $pV^{1.4} = \mathrm{const}$ gives $p_2 = \mathbf{527.8}\ \mathrm{kPa}$ and $W = (p_2 V_2 - p_1 V_1)/(1 - \gamma) = \mathbf{-48}\ \mathrm{kJ}$ - work ON the gas, and $\Delta U = -W = +48\ \mathrm{kJ}$ closes the ledger.
State function or path quantity?
Sort each quantity by whether the endpoints or the route fix it.
Isobaric work 200 kPa, 0.3 to 0.9 m3 (one decimal, kJ)?
Сложность 2/5 · 140 монет + небольшой бонус за первое завершение сегодня
W = p dV.
Isothermal work over the same tripling (one decimal, kJ)?
Сложность 2/5 · 140 монет + небольшой бонус за первое завершение сегодня
p1 V1 ln 3.
Difference between the two works (one decimal, kJ)?
Сложность 2/5 · 140 монет + небольшой бонус за первое завершение сегодня
120 - 65.9 - the area between the curves.
Adiabatic compression to 0.15 m3: final pressure (one decimal, kPa)?
Сложность 2/5 · 140 монет + небольшой бонус за первое завершение сегодня
200 x (0.3/0.15)^1.4.
Same adiabatic compression: change in internal energy (kJ)?
Сложность 2/5 · 140 монет + небольшой бонус за первое завершение сегодня
dU = -W with W = -48.
Sign and relation slips
- Work by vs work on. One sign convention, kept everywhere: $\Delta U = Q - W$ with by-positive.
- Unit mismatch in $p_1 V_1 \approx mRT$. $V$ in $\mathrm{m^3}$ with $R$ in $\mathrm{kJ}$, or the joules vanish by three decades.
- Thinking $\Delta U = m c_v \Delta T$ needs constant volume. For an ideal gas it is a state function 态函数 relation and holds on every path.
dU = m c_v dT holds for an ideal gas only in constant-volume processes.
Сложность 1/5 · 100 монет + небольшой бонус за первое завершение сегодня
It is a property relation and holds on every ideal-gas path.
Recap
- $\Delta U = Q - W$; boundary work is the area under the p-V path.
- $120$ vs $65.9$ vs $-48\ \mathrm{kJ}$: same start, three paths, three works - only $U$ (and $H$, $S$) are state functions.
- Adiabatic: $pV^{\gamma}$ constant, $\Delta U = -W$.