Q.i. Comment on the statement: no work is involved in an expansion of gas in vacuum.

$4SO_2+2O_2\rightarrow4SO_3$: $\Delta n_g = 4-6=-2$.

Balanced equation for oxidation of 4 mol $SO_2$: $4SO_2(g) + 2O_2(g) \rightarrow 4SO_3(g)$

$\Delta n_g = (\text{moles of gaseous products}) - (\text{moles of gaseous reactants})$

$\Delta n_g = 4 - (4+2) = 4 - 6 = -2$

Work of expansion is negative ($w=-p\Delta V$, $\Delta V>0$): the system does work on the surroundings.

By the standard thermodynamic sign convention used in this course, $w=-p_{ext}\Delta V$.

During expansion the gas volume increases, so $\Delta V = V_2-V_1 > 0$, making $w$ negative. This reflects that the system (the gas) is doing work ON the surroundings, so it loses energy — hence the negative sign.

With gaseous moles decreasing by 2 in this reaction, the pressure–volume work done on the system is +4.988 kJ.

For $CH_4(g) + 2O_2(g) \rightarrow CO_2(g) + 2H_2O(l)$, only $CH_4, O_2, CO_2$ are gases (water is liquid). Moles of gas: reactants = 1+2 = 3, products = 1, so

$\Delta n_g = 1 - 3 = -2$

The pressure–volume work (IUPAC convention, work done on the system) is

$w = -\Delta n_g RT = -(-2)(8.314\ JK^{-1}mol^{-1})(300\ K) = 4988.4\ J = 4.988\ kJ$

Since the gas moles contract, the surroundings compress the system, doing positive work on it.