[Needaask]

When water boils at 100 degrees and at 1 atm, ΔG=0. But at room temperature, the ΔG should be positive. However it still occurs spontaneously. I read this link http://www.vias.org/genchem/equilib_2ndlaw_12593_07.html . They said that the gas produced at produced at a lower temperature has a lower pressure and thus greater volume. So the ΔS is greater as the volume is greater so the number of micro states is greater. I have 2 questions regarding this explanation.

1) since PV=nRT, V=nRT/P so now at a higher temperature, the pressure is greater. While at lower temperatures, the pressure is smaller. So why would the lower pressure gas have a larger volume? Isn't the temperature factor not constant here?

2) even if the volume of the low pressure gas is greater, since entropy of the gas is also affected by temperature, at greater temperatures isn't the entropy of the gas also greater? Why would volume be more important here?

Thanks for the help

[Needaask]

When water boils at 100 degrees and at 1 atm, ΔG=0. But at room temperature, the ΔG should be positive. However it still occurs spontaneously. I read this link http://www.vias.org/genchem/equilib_2ndlaw_12593_07.html . They said that the gas produced at produced at a lower temperature has a lower pressure and thus greater volume. So the ΔS is greater as the volume is greater so the number of micro states is greater. I have 3 questions regarding this explanation.

1) How is ΔS calculated? The temperature is fixed so would the ΔS be the entropy of the gas at 25 degrees-entropy of liquid at 25 degrees? If so, how can the entropy of gas at 25 degrees be found since at 1 atm, the only liquid water can exist.

2) since PV=nRT, V=nRT/P so now at a higher temperature, the pressure is greater. While at lower temperatures, the pressure is smaller. So why would the lower pressure gas have a larger volume? Isn't the temperature factor not constant here?

3) even if the volume of the low pressure gas is greater, since entropy of the gas is also affected by temperature, at greater temperatures isn't the entropy of the gas also greater? Why would volume be more important here?


Thanks for the help

[curiouscat]

When water boils at 100 degrees and at 1 atm, ΔG=0. But at room temperature, the ΔG should be positive. However it still occurs spontaneously.

Huh? Water boils spontanously at room temperature? (@P=1atm)

[Needaask]

When water boils at 100 degrees and at 1 atm, ΔG=0. But at room temperature, the ΔG should be positive. However it still occurs spontaneously.

Huh? Water boils spontanously at room temperature? (@P=1atm)

Sorry I meant to say water boils at 100 degrees and with a external P of 1 atm and ΔG=0. So when water evaporates shouldn't ΔG be positive?

But I'm not too sure what the link meant. They explained that the vapour pressure is smaller so it occupies a larger volume so the entropy increase is greater. However, I was thinking the temperature of the gas now is only 25 degrees so why would the volume be greater?

And I'm not sure how they calculate the entropy change at 25 degrees since water vapour cannot exist at 25 degrees right?

Thanks

[curiouscat]

When water boils at 100 degrees and at 1 atm, ΔG=0. But at room temperature, the ΔG should be positive. However it still occurs spontaneously.

Huh? Water boils spontanously at room temperature? (@P=1atm)

Sorry I meant to say water boils at 100 degrees and with a external P of 1 atm and ΔG=0. So when water evaporates shouldn't ΔG be positive?

But I'm not too sure what the link meant. They explained that the vapour pressure is smaller so it occupies a larger volume so the entropy increase is greater. However, I was thinking the temperature of the gas now is only 25 degrees so why would the volume be greater?

And I'm not sure how they calculate the entropy change at 25 degrees since water vapour cannot exist at 25 degrees right?

Thanks

Sorry, I cannot understand what's your question here.

[Needaask]

When water boils at 100 degrees and at 1 atm, ΔG=0. But at room temperature, the ΔG should be positive. However it still occurs spontaneously.

Huh? Water boils spontanously at room temperature? (@P=1atm)

Sorry I meant to say water boils at 100 degrees and with a external P of 1 atm and ΔG=0. So when water evaporates shouldn't ΔG be positive?

But I'm not too sure what the link meant. They explained that the vapour pressure is smaller so it occupies a larger volume so the entropy increase is greater. However, I was thinking the temperature of the gas now is only 25 degrees so why would the volume be greater?

And I'm not sure how they calculate the entropy change at 25 degrees since water vapour cannot exist at 25 degrees right?

Thanks

Sorry, I cannot understand what's your question here.

Sorry for being vague.

I was wondering how we calculate the entropy change for the evaporation of water. Because the gas would be at room temperature (298K). And I don't think this is possible since gaseous water would have to be at 373K. So how do we get the entropy change such that ΔG<0?

Also, why would the ΔG of boiling or evaporation eventually reach zero? Say if we boil water at 100 degrees, ΔG is said to be 0 but isn't more and more water vaporizing? So how can my ΔG be 0? Same goes for condensation at 0 degrees where eventually all the gas turns into the liquid, how can my ΔG be 0?

Thanks so much

[Corribus]

The system is open.  A reaction can be very endergonic (non-spontaneous), but you can drive the formation of products by continually removing them from the system.  La Chatelier's Principle.

[Yggdrasil]

An important consideration here is the difference between ΔG and ΔG^o.  ΔG^o is the difference in free energy between the products and reactants in their standard state.  What does that mean?  In this case, it means the free energy of water vapor at 1 atm partial pressure and the free energy of pure liquid water.  At temperatures less that 100^oC, liquid water has the lower free energy, so water vapor cannot exist at its standard state (1 atm partial pressure).

The "real" ΔG of a process, however, takes into account the relative concentrations of products and reactants.  This quantity is defined as ΔG = ΔG^o + RT ln(Q), where Q the reaction quotient describes the concentration of reactants and products relative to their standard states.  In this case, Q would essentially just be the partial pressure of water vapor since the activity of pure water does not change in the reaction.

By setting ΔG = 0, we can find the partial pressure of water at equilibrium simply by knowing ΔG^o.  This equation is one you've probably seen before: K = e^-ΔGo/RT or ΔG^o = -RT ln(K) where K the equilibrium constant is the reaction quotient at equilibrium.  This equation tells us that at temperatures below the boiling point of water, there is some non-zero equilibrium partial pressure of water vapor.  This equilibrium partial pressure is called the vapor pressure.  If the partial pressure of water above a beaker of water is below that value, evaporation will occur spontaneously (and will continue to occur until the partial pressure of water above the beaker reaches the vapor pressure).  This is why evaporation can happen below the boiling point of water.

Essentially, the extra RT ln(Q) term when calculating the ΔG of a process occurring with products and reactants at non-standard concentrations takes into account the entropy associated with partitioning atoms between the products and reactants.  If all the atoms are only in one phase, the entropy is very low whereas having an equal amount of particles in each phase gives higher entropy.  It's this balance between the entropy associated with partitioning the atoms between the two phases and the intrinsic free energy change that comes from the liquid-to-gas transition (represented by ΔG^o) that sets the vapor pressure of water.

[Needaask]

An important consideration here is the difference between ΔG and ΔG^o.  ΔG^o is the difference in free energy between the products and reactants in their standard state.  What does that mean?  In this case, it means the free energy of water vapor at 1 atm partial pressure and the free energy of pure liquid water.  At temperatures less that 100^oC, liquid water has the lower free energy, so water vapor cannot exist at its standard state (1 atm partial pressure).

The "real" ΔG of a process, however, takes into account the relative concentrations of products and reactants.  This quantity is defined as ΔG = ΔG^o + RT ln(Q), where Q the reaction quotient describes the concentration of reactants and products relative to their standard states.  In this case, Q would essentially just be the partial pressure of water vapor since the activity of pure water does not change in the reaction.

By setting ΔG = 0, we can find the partial pressure of water at equilibrium simply by knowing ΔG^o.  This equation is one you've probably seen before: K = e^-ΔGo/RT or ΔG^o = -RT ln(K) where K the equilibrium constant is the reaction quotient at equilibrium.  This equation tells us that at temperatures below the boiling point of water, there is some non-zero equilibrium partial pressure of water vapor.  This equilibrium partial pressure is called the vapor pressure.  If the partial pressure of water above a beaker of water is below that value, evaporation will occur spontaneously (and will continue to occur until the partial pressure of water above the beaker reaches the vapor pressure).  This is why evaporation can happen below the boiling point of water.

Essentially, the extra RT ln(Q) term when calculating the ΔG of a process occurring with products and reactants at non-standard concentrations takes into account the entropy associated with partitioning atoms between the products and reactants.  If all the atoms are only in one phase, the entropy is very low whereas having an equal amount of particles in each phase gives higher entropy.  It's this balance between the entropy associated with partitioning the atoms between the two phases and the intrinsic free energy change that comes from the liquid-to-gas transition (represented by ΔG^o) that sets the vapor pressure of water.

Hi Yggdrasil sorry for the late reply

I think I understand the ΔG=ΔG°+RTlnQ reasoning where the ΔG would go to 0 and Q=k where equilibrium is reached.

But I don't quite understand the explanation if I use ΔG=G_products - G_reactants now my products is gaseous water at 298K which I don't think is possible as gaseous water is at 373K while my reactants is water at 298K.

Then for ΔG°=G°_products - G°_reactants so for the products it's 1 mole of gaseous water at 298K which also can't exist right?

So I'm having some trouble understanding this actually. Thanks for the help