Work and derivations

[JamesAntonios 1E]

So I know for a reaction that produces a net number of moles, w=-(delta n)RT, is this in anyway related to Delta H=\Delta U+\Delta nRT (which is found in the book)? If so, how come we can use this equation under constant volume. I thought that work is zero for constant volume. However, this is the problem for example 8.8. A constant-volume calorimeter showed that the heat generated by the combustion of 1.000 mol glucose molecules in the reaction C6H12O6(s)+6 O2(g)-->6 CO2(g)+6 H2O(g) is 2559 kJ at 298 K, and so U=-2559 kJ. What is the change in enthalpy for the same reaction?
and the above equation is used

[Michaela Capps 1l]

For calculating delta n (change in moles) we look at gas phase molecules in the chem equation because those are the only ones that will affect work done. In this problem, were given 6mol H20 and 6 mol CO2 in gas phase on products, totaling 12 moles, and 6 mol O2 on gas phase on the reactants (we ignore C6H12O6 because it's a solid). Product moles - Reactant moles = Delta n = 12 - 6 = 6

[804899546]

To clarify, if volume is constant but Pext is changing, would work still be done by the system? Such as if your lungs had 3L of air in them at 1 atm, then after you dive down under water the pressure is 2 atm but your lungs still remain at 3L. Could we just use w=-nRTln(P1/P2) to find work, or is it that when volume is constant work is always zero?