Answer : The value of [tex]\Delta E[/tex] for the burning of the fuel is, -998.5 joules.
Explanation :
First we have to calculate the work done.
Formula used :
[tex]w=-p_{ext}dV\\\\w=-p_{ext}(V_2-V_1)[/tex]
where,
w = work done = ?
[tex]p_{ext}[/tex] = external pressure = 1.02 atm
[tex]V_1[/tex] = initial volume of gas = 0.255 L
[tex]V_2[/tex] = final volume of gas = 1.45 L
Now put all the given values in the above formula, we get :
[tex]w=-p_{ext}(V_2-V_1)[/tex]
[tex]w=-(1.02atm)\times (1.45-0.255)L[/tex]
[tex]w=-1.22L.atm=-1.22\times 101.3J=-123.5J[/tex]
conversion used : (1 L.atm = 101.3 J)
Now we have to calculate the change in internal energy of the system.
[tex]\Delta E=q+w[/tex]
[tex]\Delta E=-875J+(-123.5J)[/tex]
[tex]\Delta E=-998.5J[/tex]
Therefore, the value of [tex]\Delta E[/tex] for the burning of the fuel is, -998.5 joules.