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These are the following statements:

[...] the thermal decomposition of $1.0$ mole $\ce {CaCO3}(\text{s})$ at $1$ bar results in an increase of volume of nearly $90~\text{dm}^3$ at $800^\circ \text{C}$ on account of the carbon dioxide produced.

Is the system at $1$ bar or the surrounding exerts pressure of $1$ bar? Is $800^\circ \text{C}$ the temperature of the system or the surroundings?

The standard state of a pure substance is the pure substance exactly at $1$ bar & at $298.15~\text K$.

I think the pressure mentioned is of surroundings but the temperature? Is it of the system or he surroundings?

We see that at constant temperature & pressure, the change in Gibbs energy of a system is proportional to the overall change in entropy of the system plus its surroundings.

Now is the system at constant temperature & pressure or the surroundings?

Could anyone please explain me to clear this ambiguity?

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When we say

$$\Delta G^\circ = \Delta H^\circ - T\Delta S^\circ$$

we are referring to $\Delta G^\circ$, $\Delta H^\circ$ and $\Delta S^\circ$ of the system. As such, the variable $T$ in the equation refers to the temperature of the system. The condition for constant pressure $p$ is implicit; the equation above does not directly show why $p$ has to be constant. For more information, refer to this question and answer.

The sentences that you have given are not the best examples to explain this, because they do not relate to Gibbs free energy. I will give you a different example:

When hydrochloric acid is added to sodium carbonate at atmospheric pressure, the standard enthalpy change $\Delta H^\circ$ and the standard entropy change $\Delta S^\circ$ are measured to be $x~\mathrm{kJ~mol^{-1}}$ and $y~\mathrm{J~K^{-1}~mol^{-1}}$ respectively. Calculate the Gibbs free energy of the reaction and hence determine the equilibrium constant. (The temperature at which the reaction is carried out is $25~\mathrm{^\circ C}$.)

The wording of the question may seem to imply that it is talking about the temperature and pressure of the surroundings, and that is probably not a coincidence. For a simple chemical reaction like the one above, we would probably carry it out in a beaker. We weigh out a certain mass of $\ce{Na2CO3}$, and add a certain volume of $\ce{HCl}$, and pour the acid into the beaker and watch the bubbles.

Under such conditions, we can speak of the system being at mechanical and thermal equilibrium. Without going into too much detail about reversible and irreversible processes (I know there are loopholes in the explanation):

  • Mechanical equilibrium, in general, means that the net force acting on the system is zero. Correspondingly, this must mean that the pressure of the system and the pressure of the surroundings are equal at all points in time. If, at any point in time, this was not the case, e.g. if we had $p_\mathrm{syst} < p_\mathrm{surr}$, then the surroundings would "push" on the system and cause the system to contract until it reaches the point where $p_\mathrm{syst} = p_\mathrm{surr}$.

  • Thermal equilibrium means that there is no heat transfer into, or out of, the system. Analogously to above, this must mean that the temperature of the system and the temperature of the surroundings are equal at all points in time. We conventionally assume the surroundings to be a "reservoir" which maintains a constant temperature, and this is pretty true as far as our simple example is concerned. Therefore $T_\mathrm{syst} = T_\mathrm{surr}$.

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  • $\begingroup$ Does that mean all processes need to be isothermal & isobaric? The standard enthalpy of ionization of hydrogen is +1312 kJ . Does that mean this energy is to be supplied to ionize hydrogen which is at 1 bar & 298.15 K? Does that mean during the process, the temperature must remain same? If so, then how could even there be exchange of heat, if the hydrogen remained at the same temperature as that of the surroundings all the time during the process? Gibbs Free energy is applicable when the process is isothermal-isobaric. Is it so? This is what I'm not understanding. [contd.] $\endgroup$
    – user5764
    Nov 28, 2015 at 3:56
  • $\begingroup$ If the system & surroundings are at same temperature 298.15 K throughout the process, how could there be supply of heat after-all this requires temperature-gradient, isn't it? Do every chemical reactions happen in the same way where throughout the process, the system & surroundings remain at same temperature? Can you please clarify that? $\endgroup$
    – user5764
    Nov 28, 2015 at 3:57
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Your first two statements mean that, if T and p are the temperature and pressure of your system in its initial thermodynamic equilibrium state, the surroundings will be maintained at the temperature T and the pressure p throughout the change from the initial thermodynamic equilibrium state to the final thermodynamic equilibrium state (say by using a constant temperature reservoir and an external gas pressure). Thus, in the final thermodynamic equilibrium state of the system, the temperature and pressure will also be T and p. However, during the change between the initial and final thermodynamic equilibrium states, the temperatures and pressures within the system can vary spatially and with time. Only at the boundary with the surroundings will the temperature and pressure match that of the surroundings throughout the process. For more details, see Section 2.4 Properties of the Gibbs Function, in Denbigh, Principles of Chemical Equilibrium.

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  • $\begingroup$ So, throughout the process, the substances needs to be at the same temperature as that of the surroundings, ?I s it so? There should be no increase or decrease in temperature of the substances throughout the process. Is it so? But how could there be exchange of heat if the system & surroundings all remain at the same temperature throughout the process? There needs to be a temperature-gradient. Isn't it? Also, if heat is being supplied or taken away from the system, wouldn't its temperature increase or decrease & thus remaining at the same temperature as that of the surrounding? Confused. $\endgroup$
    – user5764
    Nov 28, 2015 at 4:04
  • $\begingroup$ What do the words "However, during the change between the initial and final thermodynamic equilibrium states, the temperatures and pressures within the system can vary spatially and with time" mean to you? $\endgroup$
    – Chet Miller
    Nov 28, 2015 at 12:38
  • $\begingroup$ So, during the process, the temperature & pressure within the system doesn't remain uniform? I thought Gibbs energy is applicable only for isothermal-isobaric process. Also, the equation $$G= H- TS$$ is a state function which implies $T$ is the temperature of the system, isn't it? So, $$\Delta G = \Delta H-T\Delta S $$ doesn't imply the change in entropy at constant temperature $T$ of the state? $\endgroup$
    – user5764
    Nov 28, 2015 at 12:54
  • $\begingroup$ The statements I was addressing didn't say anything about the thermodynamic functions. But, irrespective of how the process is carried out, the change in the thermodynamic functions depends only on the initial and final thermodynamic equilibrium states. So the Gibbs energy doesn't only apply to isothermal-isobaric processes. For a process in which the pressure in the system doesn't remain uniform, even though it is held constant at the boundary, q is still equal to $\Delta H$. $\endgroup$
    – Chet Miller
    Nov 28, 2015 at 14:52
  • $\begingroup$ Yes, they are state-functions. But could you tell why the site I linked wrote isothermal-isobaric system? $\endgroup$
    – user5764
    Nov 28, 2015 at 15:38
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It is important think of the thermodynamic functions U, H, S, G, and A as physical properties of the material(s) comprising the system, not determined by the path of any particular process. But note that, since the only way we have of evaluating the change in S between two thermodynamic equilibrium states of a system is to calculate the integral of dq/T for a reversible path. This is independent of the actual path that took the system from its initial state to its final state. Since G and A are also functions of S, the only way we have of evaluating the changes G and A are also to determine them for a reversible path. This too is also independent of the actual path that took the system from its initial state to its final state.

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