Heat of combustion of ethanol at constant pressue and at temperature `TK(=298K)` is found to be `-qJmal^(-1)` .
Hence, heat of combustion (in `Jmol^(-1)`) of ethanol at the same temperature at constant volume will be:

Options

A. `RT-q`
B. `-(q+RT)`
C. `q-RT`
D. `q+RT`

Answer

C

Explanation

`C_(2)H_(5)OH(l)+3O_(2)(g)rarr2CO_(2)(g)+3H_(2)O(l)`
`Deltan_(g)=2-3=-1`
so `DeltaU=DeltaH-Deltan_(g)RT`
`=-q+RT`