恒温压缩热力学计算

생성일: 2025년 1월 19일

답변에 사용된 모델 O1 by Chat01

질문

1 mol 理想气体，从 298 K，100 kPa，经恒温，恒外压为 600 kPa 时，压缩至终态。
计算过程的 Q、W、ΔU、ΔH、ΔS、ΔA、ΔG。

사고

Isothermal compression challenge

I'm grappling with isothermal compression where the external pressure is 600 kPa, higher than the gas's initial 100 kPa. This seems a unique approach.

Heat, work, and pressure

Healing or working through the change involves calculating heat, work, and various energy shifts within the system, especially with the final pressure set at 600 kPa.

Tracking volume shifts

I’m piecing together the initial and final states of the ideal gas. The initial volume is 24.8 L, and after compression to 600 kPa, it reduces to 4.13 L.

Breaking down isothermal processes

In this irreversible isothermal compression, $\Delta U$ and $\Delta H$ are zero because temperature and enthalpy don't change for an ideal gas, despite external pressure shifts.

Breaking down work dynamics

OK, let me see. In isothermal compression, the work done on the gas is $W = - P_{\text{ext}} (V_f - V_i)$ , with the sign indicating the system's work perspective.

Evaluating work direction

For irreversible compression with constant external pressure, the work done by the system is determined as $W = P_{\text{ext}} (V_f - V_i)$ , which ends up negative.

Balancing work

The definition of work when the system is compressed suggests positive work, but maintaining consistent sign conventions reveals the system actually does negative work as volume decreases.

Evaluating work dynamics

Interestingly enough, the work done on the system (compression) is negative because the system loses energy, while work done by the system would mean it gains energy.

Clarifying work dynamics

OK, let me see: if the volume decreases, the net work done is positive, indicating the system gains energy from the surroundings.

Clarifying sign convention

The problem's sign convention aligns with Chinese thermodynamics, where the work done on the system during compression is positive, emphasizing energy interaction with the surroundings.

Defining work dynamics

In Chinese thermodynamics, $W = P_{\text{ext}} (V_i - V_f)$ is positive, indicating work done on the system during compression, clarifying energy transfer with the surroundings.

Balancing energy transfer

We can see that $W = P_{\text{ext}} (V_i - V_f)$ is correct for compression. If $W$ is positive, the system loses heat to its surroundings, aligning with isothermal compression of an ideal gas.

Calculating precise work

I'm moving towards quantifying $W$ using specific pressures and temperatures. Once $W$ is established, I can find $Q$ and ensure clarity in units and final energy expressions.

Calculating initial volume

I’m working through the initial volume calculation using standard pressure and temperature, which comes to 24.76 liters. This provides a baseline for further volume and energy calculations.

Calculating work done

I’ve pinpointed the work done during isothermal compression, arriving at 12.38 kJ by multiplying the external pressure by the volume difference.

Releasing energy

I’m confirming $Q = -12.38$ kJ, aligning with isothermal compression, where the system releases 12.38 kJ to the surroundings, and $W = +12.38$ kJ.

Assessing entropy shifts

I’m calculating the entropy change for an isothermal ideal gas process. The calculated $\Delta S$ is approximately -14.9 J/K, revealing how entropy decreases as the gas is compressed.

Expressing entropy differences

OK, let me see. Using $\Delta S$ in J/K for 1 mol, I’m calculating the system's entropy change as the gas undergoes isothermal compression.

Verifying entropy consistency

OK, let me see. The entropy change $\Delta S$ accurately measures compression at 1 mol, confirming it as -14.9 J/K for the system.

Calculating Helmholtz free energy

I’m working through the Helmholtz free energy change $\Delta A$ by using the entropy change $\Delta S$ and the constant temperature of 298 K.

Calculating Gibbs free energy

I’m working out the Gibbs free energy change $\Delta G$ for an isothermal process by combining the enthalpy change $\Delta H$ and the entropy change $\Delta S$ .

Refining energy terms

I’m compiling $Q$ , $W$ , $\Delta U$ , $\Delta H$ , $\Delta S$ , $\Delta A$ , and $\Delta G$ in consistent kJ and J/K units, emphasizing clarity and coherence.

Refining calculations

I’m finalizing the energy term calculations, ensuring clarity by using consistent units and significant figures throughout. This helps solidify a comprehensive understanding of the process.

Thought about ideal gas thermodynamics，持续 1m 39s

답변

下面给出一种典型的解题思路和结果汇总。题意是：1 mol 理想气体，初态 $T=298\,\mathrm{K}$ 、 $p=100\,\mathrm{kPa}$ ，在恒温条件下对外作业（外压恒为 $600\,\mathrm{kPa}$ ），将气体压缩到终态。要求计算该不可逆恒温过程中下列量的数值：

Q,\quad W,\quad \Delta U,\quad \Delta H,\quad \Delta S,\quad \Delta A,\quad \Delta G.

1. 初、终态参数

物质的量： $n = 1\,\mathrm{mol}$
气体常数： $R = 8.314\,\mathrm{J\,(mol\cdot K)^{-1}}$
初态：
$T_i = 298\,\mathrm{K},\quad p_i = 100\,\mathrm{kPa} = 1.00\times 10^5\,\mathrm{Pa}.$
理想气体状态方程 $pV = nRT$ ，可得初态体积
$V_i = \frac{n\,R\,T_i}{p_i} = \frac{1\times 8.314\times 298}{1.00\times 10^5} \approx 0.02476\,\mathrm{m^3} \approx 24.76\,\mathrm{L}.$
终态：
题目暗示最终与外界平衡时，气体压力变为 $p_f = 600\,\mathrm{kPa} = 6.00\times 10^5\,\mathrm{Pa}$ ，温度仍为 $298\,\mathrm{K}$ 。
因为仍为理想气体，终态体积
$V_f = \frac{n\,R\,T}{p_f} = \frac{1\times 8.314\times 298}{6.00\times 10^5} \approx 0.00413\,\mathrm{m^3} \approx 4.13\,\mathrm{L}.$

2. 过程分析与各热力学量的计算

2.1 $\Delta U$ 和 $\Delta H$

对于理想气体，内能 $U$ 和焓 $H$ 都只与温度有关。当过程是恒温时，

\Delta U = 0, \quad \Delta H = 0.

2.2 体积功 $W$

若采用热力学常用符号约定( $\Delta U = Q + W$ )，则 $W$ 为外界对系统所做的功。
在不可逆恒压(对外压 $p_\mathrm{ext} = 600\,\mathrm{kPa}$ 不变)压缩下，外界对气体做功

W = p_\mathrm{ext}\,\bigl(V_i - V_f\bigr).

注意这里体积减少，因此 $(V_i - V_f) > 0$ ，数值上：

p_\mathrm{ext} = 6.00\times 10^5\,\mathrm{Pa},\quad V_i - V_f \approx 0.02476 - 0.00413 = 0.02063\,\mathrm{m^3}.

所以

W = 6.00\times 10^5\,\mathrm{Pa}\times 0.02063\,\mathrm{m^3} \approx 1.24\times 10^4\,\mathrm{J} = 12.4\,\mathrm{kJ}.

2.3 吸放热量 $Q$

由第一定律

\Delta U \;=\; Q + W.

由于 $\Delta U = 0$ ，可得

Q = -\,W = -\,12.4\,\mathrm{kJ}.

这表示在不可逆恒温压缩过程中，气体放热 12.4 kJ。

2.4 熵变 $\Delta S$

熵是状态函数，故气体的熵变可用初、终态直接计算；对于 1 mol 理想气体的等温过程，

\Delta S = n\,R\,\ln\!\Bigl(\frac{V_f}{V_i}\Bigr) = R\,\ln\!\Bigl(\frac{p_i}{p_f}\Bigr) \;(\text{等温时 }pV=\text{常数}) .

带入数值：

\frac{p_i}{p_f} = \frac{100\,\mathrm{kPa}}{600\,\mathrm{kPa}} = \frac{1}{6}, \quad \ln\!\Bigl(\frac{1}{6}\Bigr) = -\ln(6) \approx -1.79176.

\Delta S = 8.314\times(-1.79176) \approx -14.9\,\mathrm{J\,K^{-1}}.

这里得到的熵变为负值，说明在此不可逆恒温压缩过程中，气体自身的熵减少。

2.5 $\Delta A$ (Helmholtz 自由能变)

Helmholtz 自由能定义 $A = U - TS$ 。若温度恒定，则

\Delta A = \Delta U \;-\; T\,\Delta S.

因为 $\Delta U = 0$ ，所以

\Delta A = -\,T \,\Delta S = -\,298\,\mathrm{K}\,\times\bigl(-14.9\,\mathrm{J\,K^{-1}}\bigr) \approx +\,4440\,\mathrm{J} = +\,4.44\,\mathrm{kJ}.

2.6 $\Delta G$ (Gibbs 自由能变)

对于理想气体， $\Delta H=0$ 。而吉布斯自由能定义 $G = H - TS$ 。也可用常见公式

\Delta G = \Delta H \;-\; T\,\Delta S.

因为 $\Delta H=0$ ，故

\Delta G = -\,T\,\Delta S = +\,4.44\,\mathrm{kJ}.

3. 结果汇总

$\displaystyle \Delta U = 0$
$\displaystyle \Delta H = 0$
$\displaystyle W = +12.4\,\mathrm{kJ}$
$\displaystyle Q = -12.4\,\mathrm{kJ}$
$\displaystyle \Delta S = -14.9\,\mathrm{J\,K^{-1}}$
$\displaystyle \Delta A = +4.44\,\mathrm{kJ}$
$\displaystyle \Delta G = +4.44\,\mathrm{kJ}$

以上 $W$ 的正号表示外界对系统做功

$Q<0$ 表示系统向外放热

$\Delta S<0$ 表示气体熵减

$\Delta A,\Delta G>0$ 表示该过程使体系的自由能升高（因这是一个不可逆自外界推动的压缩过程）。

질문

사고

답변

1. 初、终态参数

2. 过程分析与各热力学量的计算

2.1 ΔU\Delta UΔU 和 ΔH\Delta HΔH

2.2 体积功 WWW

2.3 吸放热量 QQQ

2.4 熵变 ΔS\Delta SΔS

2.5 ΔA\Delta AΔA (Helmholtz 自由能变)

2.6 ΔG\Delta GΔG (Gibbs 自由能变)