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) $H_{2}O{(l,1 bar, 273K)}$ $\rightarrow$$H_{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!

  • 1
    $\begingroup$ 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. $\endgroup$
    – Poutnik
    Commented Apr 1, 2023 at 5:42
  • $\begingroup$ Have you tried Delta S_surr = integral ( -dQ_rev,sys / T) ? That for 3 formal stages: heating up, evaporation, cooling down. $\endgroup$
    – Poutnik
    Commented Apr 1, 2023 at 7:46
  • $\begingroup$ @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. $\endgroup$
    – Natasha J
    Commented Apr 1, 2023 at 8:14
  • 1
    $\begingroup$ 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) $\endgroup$
    – Poutnik
    Commented Apr 1, 2023 at 8:55
  • $\begingroup$ Then what should I use for dQ? dQ= the values of enthalpy that were calculated before? $\endgroup$
    – Natasha J
    Commented Apr 1, 2023 at 9:52

1 Answer 1


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 $\pu{100 K}$.
  • $\begingroup$ For readers, VLE stands for Vapor Liquid Equilibrium $\endgroup$
    – Poutnik
    Commented Apr 2, 2023 at 17:10
  • $\begingroup$ Thank you so much for explaining. I was stuck at it for hours.. Thanks again $\endgroup$
    – Natasha J
    Commented Apr 3, 2023 at 2:13

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