4D.7 7th Edition

[vuongnaomi1L]

(a) Calculate the work that must be done against the atmosphere for the expansion of the gaseous products in the combustion of
1.00 mol C6H6(l) at 25°C and 1.00 bar. (b) Using data in Appendix 2A, calculate the standard enthalpy of the reaction. (c) Calculate the change in internal energy, ΔU, of the system.

so I understand for part a, we have to use w=-P∆V=-nRT, but I am confused on why for n we use -1.5 moles, which is the moles of gas of the products minus the moles of gas for the reactants in -nRT.

[Carolina 3E]

DeltaV is related to deltan
PdeltaV=deltanRT
In the balanced chemical equation the difference but the moles of gas of products and the moles of gas of reactants is -1.5 moles

[edward_brodell_2I]

1 C6H6(l) + 7.5 O2(g) -> 6 CO2(g) and 3 H2O(g)
Change in moles of gas is nproducts - nreactants => 9 mol (g) - 7.5 mol (g) = 1.5 mol of gas
PV=nRT and since pressure is constant, the change in volume is the same as the net change in the number of moles of gas brought about by the reaction.