[Needaask]

ΔH=ΔE+PΔV So if there is an expansion in the system the work done must be taken into account. However, I don't understand why the external pressure must be fixed. Would the reason why it must be fixed be because of the PΔV part? where if it changes suddenly then so will the pressure exerted? I'm not too sure if this is the reason because that would mean for reactions with no expansion, the pressure doesn't have to be fixed. But I don't think that is true. So actually why must the pressure be fixed?

Also, the Gibbs Free Energy Formula is this: ΔG=ΔH-TΔS where the temperature is constant. I have some questions regarding why T is constant.
1) Since ΔH is present, heat will be absorbed/given out so won't there be a temperature change?
2) For ΔS one of the factors affecting it is temperature (Higher T=Higher entropy level) and when we raise the temperature the ΔS increases. And again the temperature increases from the initial state to the final state. So how can T be a constant?

[curiouscat]

ΔH=ΔE+PΔV So if there is an expansion in the system the work done must be taken into account. However, I don't understand why the external pressure must be fixed. Would the reason why it must be fixed be because of the PΔV part? where if it changes suddenly then so will the pressure exerted? I'm not too sure if this is the reason because that would mean for reactions with no expansion, the pressure doesn't have to be fixed. But I don't think that is true. So actually why must the pressure be fixed?

No reason for P to be fixed. If P varies your way to calculate work will be more involved, that's all.

[Corribus]

Also, the Gibbs Free Energy Formula is this: ΔG=ΔH-TΔS where the temperature is constant. I have some questions regarding why T is constant.
1) Since ΔH is present, heat will be absorbed/given out so won't there be a temperature change?

In a closed, system, yes.  But we assume there is a reservoir for heat exchange.  In such a case, endothermicity and exothermicity of a reaction effectively don't result in a temperature change, assuming heat conduction is sufficiently fast.  Bear in mind, Gibbs energy is something of an idealization.  For most chemical reactions, the approximations are such that the Gibbs energy can give a good estimation of the thermodynamics of the chemical change.  However in some cases - such as explosion chemistry - where temperature and pressure changes overcome the surroundings' capacity to compensate (and approximate as "constant temperature or pressure"), Gibbs energy is no longer so useful.

2) For ΔS one of the factors affecting it is temperature (Higher T=Higher entropy level) and when we raise the temperature the ΔS increases. And again the temperature increases from the initial state to the final state. So how can T be a constant?

Pretty much the same as above, although the temperature impact of entropy is taken into account in the Gibbs expression.  Generally, we can assume that ΔS and ΔH themselves are independent of temperature, for small temperature changes anyway.

[Needaask]

ΔH=ΔE+PΔV So if there is an expansion in the system the work done must be taken into account. However, I don't understand why the external pressure must be fixed. Would the reason why it must be fixed be because of the PΔV part? where if it changes suddenly then so will the pressure exerted? I'm not too sure if this is the reason because that would mean for reactions with no expansion, the pressure doesn't have to be fixed. But I don't think that is true. So actually why must the pressure be fixed?

No reason for P to be fixed. If P varies your way to calculate work will be more involved, that's all.

hi in what way would calculating work done be more difficult if the pressure wasn't constant?

Thanks

[Needaask]

Also, the Gibbs Free Energy Formula is this: ΔG=ΔH-TΔS where the temperature is constant. I have some questions regarding why T is constant.
1) Since ΔH is present, heat will be absorbed/given out so won't there be a temperature change?

In a closed, system, yes.  But we assume there is a reservoir for heat exchange.  In such a case, endothermicity and exothermicity of a reaction effectively don't result in a temperature change, assuming heat conduction is sufficiently fast.  Bear in mind, Gibbs energy is something of an idealization.  For most chemical reactions, the approximations are such that the Gibbs energy can give a good estimation of the thermodynamics of the chemical change.  However in some cases - such as explosion chemistry - where temperature and pressure changes overcome the surroundings' capacity to compensate (and approximate as "constant temperature or pressure"), Gibbs energy is no longer so useful.

2) For ΔS one of the factors affecting it is temperature (Higher T=Higher entropy level) and when we raise the temperature the ΔS increases. And again the temperature increases from the initial state to the final state. So how can T be a constant?

Pretty much the same as above, although the temperature impact of entropy is taken into account in the Gibbs expression.  Generally, we can assume that ΔS and ΔH themselves are independent of temperature, for small temperature changes anyway.

Oh but when calculating ΔH we would use the temperature change and multiply it with the heat capacity of the system? So I don't get how we can the conditions for ΔH or ΔG or ΔS to have a constant temperature.

Cos in so many reactions won't the temperature change already be significant? And even in in ΔS one of the factors affecting it is a change in temperature already. So I also feel like there is a contradiction here.

Thanks for the help

[curiouscat]

ΔH=ΔE+PΔV So if there is an expansion in the system the work done must be taken into account. However, I don't understand why the external pressure must be fixed. Would the reason why it must be fixed be because of the PΔV part? where if it changes suddenly then so will the pressure exerted? I'm not too sure if this is the reason because that would mean for reactions with no expansion, the pressure doesn't have to be fixed. But I don't think that is true. So actually why must the pressure be fixed?

No reason for P to be fixed. If P varies your way to calculate work will be more involved, that's all.

hi in what way would calculating work done be more difficult if the pressure wasn't constant?

Thanks

Integral. PdV