Answer
(a) $\Delta n = 0$, so the volume does not affect the equilibrium constant.
(b) $K_c = K_p = 0.659$
Work Step by Step
(a) There are 2 moles of gaseous reactants and 2 moles of gaseous products. $\Delta n = 0$, therefore, the change in volume will not affect the equilibrium.
Mathematically:
$$K_c = \frac{[CO][H_2O]}{[CO_2][H_2]}$$
Since molarity is number of moles divided by volume:
$$K_c = \frac{\frac{n_{CO}}{V}{}\frac{n_{H_2O}}{V}}{\frac{n_{CO_2}}{V}\frac{n_{H_2}}{V}}$$
Simplifying:
$$K_c = \frac{{n_{CO}}{}{}{n_{H_2O}}{}}{{n_{CO_2}}{}{n_{H_2}}{}}$$
The Volume do not appear on the equilibrium constant expression.
(b) Substitute:
$$K_c = \frac{(0.224)(0.224)}{(0.276)(0.276)} = 0.659$$
$$K_p = K_c(RT)^{\Delta n} = K_c(RT)^0 = K_c(1) = K_c = 0.659$$