[sodium.dioxid]

When calculating work done by the chemical system, why is the atmospheric pressure used instead of the pressure generated against the atmosphere by the system? Here is a picture to show you what I am talking about.



Unlike the pressure of the atmosphere, the pressure of the system is not constant. It keeps decreasing by pushing up until the two pressures are equal and opposite. So, shouldn't work equal -P_avg. x ΔV, where P_avg. = P(initial)-P(atmospheric)/2?

I am assuming the reaction is instantaneous, where p(initial) generated by the system has a maximum value.

[sodium.dioxid]

Let me put this another way. Let us say that we are helping push down by hand a piston that is waiting to decompress (like a spring). The piston would go X meters. What I don't understand is from the initial position to X/2meters (the piston having traveled half of its total distance), the expressions "w=-pv" and "w=-fd" do not give the same results (even after converting units). The problem is that the force generated by the gas is continually declining; it decreases as the piston gets pushed further. On the contrary, the pressure of the atmosphere is constant. If we use the expression "w=-pv", then the energy released from initial position to X/2 is the same as that released when the piston moves from X/2 to X. This is not the case when we use "w=-fd". The forces involved through the first half of the distance are greater than those involved in the second half are, since force is constantly decreasing as the volume gets larger. Am I making any faulty assumptions?

[Jorriss]

How are you using w=-fd to calculate the work done by moving an entire surface? dw=-pdv IS dw=-fdr for a piston expanding.

[juanrga]

When calculating work done by the chemical system, why is the atmospheric pressure used instead of the pressure generated against the atmosphere by the system? Here is a picture to show you what I am talking about.



Unlike the pressure of the atmosphere, the pressure of the system is not constant. It keeps decreasing by pushing up until the two pressures are equal and opposite. So, shouldn't work equal -P_avg. x ΔV, where P_avg. = P(initial)-P(atmospheric)/2?

I am assuming the reaction is instantaneous, where p(initial) generated by the system has a maximum value.

The work done by a system depends of the pressure of the system -p_systdV. Classical thermodynamics only deals with equilibrium states and 'quasistatic' processes and for those the pressure of the system is always equilibrated by the pressure of the surrounds. Therefore p_systdV = p_surrdV