# How do I find entropy change for surroundings?

I know how to calculate entropy change of the system but am not able to calculate entropy change for surroundings. The question is as follows:

Q: At $$100°C$$, water vapour at $$1$$ bar is in equilibrium with the liquid.

Given: $$\ce{H2O}\mathrm{(l,1\,bar, 373\,K) } \to \ce{H2O}\mathrm{ (g,1\,bar,373\, K) }$$ ;

$$\Delta_{r} H$$ $$=$$ $$40.639 kJmol^{-1}$$, $$C_{p}(g) = 30.305 JK^{-1}mol^{-1}$$, $$C_{p}(l) = 75.31 JK^{-1}mol^{-1}$$. Compute $$\Delta_{r}S$$ for the reaction: $$H_{2}O{(l,1 bar, 273K)}$$ $$\rightarrow$$ $$H_{2}O(g,1 bar,273K)$$ and also $$\Delta S_{surr}$$ and $$\Delta S_{total}$$.

I was able to calculate the entropy change of system. It came out to be, $$\Delta S_{sys} = 123JK^{-1}mol^{-1}$$. But I am stuck at change in entropy of surroundings.

I calculated the change in entropy of system by following this sequence:

A) $$H_{2}O{(l,1 bar, 273K)}$$ $$\rightarrowH_{2}O{(l,1 bar, 373K)}$$

B) $$H_{2}O{(l,1 bar, 373K)}$$ $$\rightarrow$$ $$H_{2}O{(g,1 bar, 373K)}$$

C) $$H_{2}O{(g,1 bar, 373K)}$$ $$\rightarrow$$ $$H_{2}O{(g,1 bar, 273K)}$$

Any guidance is appreciated. Thanks!

• Reference for chem equation/expression formatting: Notation basics / Formatting of math/chem expressions / upright vs italic // For more: Math SE MathJax tutorial. // Not to be applied in CH SE titles. Apr 1, 2023 at 5:42
• Have you tried Delta S_surr = integral ( -dQ_rev,sys / T) ? That for 3 formal stages: heating up, evaporation, cooling down. Apr 1, 2023 at 7:46
• @Poutnik, I tried this thing $\Delta H$ $=$ $\int_{T_{1}}^{T_{2}} C_{p,m}\ dT$ for three stages and then I divided the three $\Delta H$ by temperature 273K, 373K and 373K for three stages respectively. Is this what you were trying to say? My answer came out different than what was given in the textbook though. Apr 1, 2023 at 8:14
• You have to keep 1/T in the integrals, as T changes in 2 of 3 stages. dS=dQ/T --> Delta S=Delta Q/T is valid only at constant T. (written on phone, no fancy formatting now) Apr 1, 2023 at 8:55
• Then what should I use for dQ? dQ= the values of enthalpy that were calculated before? Apr 1, 2023 at 9:52

As Chet Miller pointed in the comments, water is not in VLE at $$P = 1 \; \pu{bar}$$ and $$T = 273 \; \pu{K}$$. Thereby the entropy calculation you made needs some correction. Lets do that first and then we answer your question. Let me also name: $$P_1 = \pu{unknown saturation pressure}$$, $$P_2 = \pu{1 bar}$$, $$T_1 = \pu{273 K}$$, and $$T_2 = \pu{373 K}$$.

Calculation of Entropy

1. We estimate the saturation pressure at $$T_1$$ by the Clausius-Clapeyron equation \begin{align} P_1 &= P_2 \exp\bigg[-\dfrac{\Delta_\pu{vap} H}{R}\bigg(\dfrac{1}{T_1} - \dfrac{1}{T_2}\bigg)\bigg] \\ P_1 &= (\pu{1 bar}) \exp\bigg[-\dfrac{\pu{40639 J/mol}}{\pu{8.314 J/mol K}}\bigg(\dfrac{1}{\pu{273 K}} - \dfrac{1}{\pu{373 K}}\bigg)\bigg] \\ P_1 &= \pu{8.230E-3 bar} = \pu{0.823 kPa} \end{align} Water tables inform that the exact value is $$\pu{0.611 kPa}$$ ($$\pu{35 \%}$$ relative error), but lets only work with the data given.

2. I will add two steps, A and E, to your thermodynamic path and take note of the entropies that you have calculated: \begin{align} &A) \ce{H2O}(P_1, T_1, \pu{liquid}) \rightarrow \ce{H2O}(P_2, T_1, \pu{liquid}) \\ &B) \ce{H2O}(P_2, T_1, \pu{liquid}) \rightarrow \ce{H2O}(P_2, T_2, \pu{liquid}) \hspace{1 cm} \Delta S_B = \pu{23.50 J/mol K} \\ &C) \ce{H2O}(P_2, T_2, \pu{liquid}) \rightarrow \ce{H2O}(P_2, T_2, \pu{gas}) \hspace{1 cm} \Delta S_C = \pu{108.95 J/mol K} \\ &D) \ce{H2O}(P_2, T_2, \pu{gas}) \rightarrow \ce{H2O}(P_2, T_1, \pu{gas}) \hspace{1 cm} \Delta S_D = \pu{-9.46 J/mol K} \\ &E) \ce{H2O}(P_2, T_1, \pu{gas}) \rightarrow \ce{H2O}(P_1, T_1, \pu{gas}) \\ \end{align} The three entropies calculated sum to that value you posted. We need two additional ones. For a constant temperature process, the entropy change is \begin{align} \Delta S = \int_{P_1}^{P_2}-\bigg(\dfrac{\partial V}{\partial T}\bigg)_P \; dP \end{align} With the data given, we cannot calculate the integrand for the liquid state (although we could try to use an EOS, if you want, tell me in the comments), so I will assume $$\Delta S_A = 0$$. For the gas state, application of the ideal gas law yields \begin{align} \Delta S_E = -R\ln\bigg(\dfrac{P_1}{P_2}\bigg) \rightarrow \Delta S_E = 39.91 \; \pu{J/mol K} \end{align}

3. Adding this to the sum yields $$\boxed{\Delta_\pu{vap} S = 162.91 \; \pu{J/mol K}}$$ Water tables state that $$\Delta_\pu{vap} S = 164.98 \; \pu{J/mol K}$$. Quite a good guess IMO considering the simplifications.

Calculation of heat to the surroundings

Lets calculate all the heat exchanged with the surroundings. Steps B-D are constant pressure processes, and we know that $$Q = \Delta H$$, so they are immediate \begin{align} Q_B = \Delta H_B = c_\pu{p,l} (T_2 - T_1) &\rightarrow Q_B = 7531 \; \pu{J/mol} \\ Q_C = \Delta H_C = \Delta_\pu{vap} H &\rightarrow Q_C = 40639 \; \pu{J/mol} \\ Q_D = \Delta H_D = c_\pu{p,g} (T_1 - T_2) &\rightarrow Q_D = -3030.5 \; \pu{J/mol} \end{align} For steps A and E, for a constant temperature process, we know that $$Q = T \Delta S$$ \begin{align} Q_A &= T_1 \Delta S_A \rightarrow Q_A = 0 \\ Q_E &= T_2 \Delta S_E \rightarrow Q_E = 14886.08 \; \pu{J/mol} \end{align} And summing all the heats we have $$Q = 60025.58 \; \pu{J/mol}$$. This value makes sense, and is the heat that the surroundings must give to the system to make possible those 5 steps.

To be fair, this is as far as we can get; for the surroundings we need to make some assumptions. The simplest one is to consider it as a heat reservoir, i.e., a system that can exchange infinite amounts of energy without changing its temperature. However, there are three restrictions:

1. The temperature of the reservoir must be at least of $$\pu{273 K}$$ for step A, or we would violate the 2nd law.
2. Likewise, the temperature of the reservoir must be at least $$\pu{373 K}$$ for step C, or we would violate the 2nd law.
3. Point 2 applies equally to step E.

The best case scenario we can postulate, in terms of reversibility, is that of a surroundings at this last temperature, and thus \begin{align} \Delta S_\pu{sys} + \Delta S_\pu{surr} \geq 0 \\ \Delta S_\pu{sys} + \dfrac{Q_\pu{surr}}{T_\pu{surr}} \geq 0 \\ 162.91 \; \dfrac{\pu{J}}{\pu{mol K}} - \dfrac{60025.85 \; \pu{J/mol}}{373 \; \pu{K}} \geq 0 \\ 1.981 \dfrac{\pu{J}}{\pu{mol K}} \geq 0 \end{align} So one possible answer is $$\boxed{\Delta S_\pu{surr} = -160.93 \; \dfrac{\pu{J}}{\pu{mol K}}}$$

1. I don't like the symbol $$\Delta_r H$$ suggested by your textbook. That it not a chemical reaction, it is a phase transition. So I would write $$\Delta_\pu{vap} H$$ instead.
2. Reversibility can be attained by refining the heat exchanges for the heating (step B) and the cooling (step D). For example, heating some amount with a reservoir of $$\pu{50 K}$$ and then the other amount with one of $$\pu{100 K}$$.