Answer
$P_{H2O}=7.90\ kPa$
$T_{cond}=41.3^{∘}C$
Work Step by Step
The total mole of the mixture and the mole fraction of water vapor are
$$
\begin{aligned}
& N_{\text {total }}=8+9+4+94=115 \mathrm{kmol} \\
& y_{\text {H2O }}=\frac{N_{\text {H12O }}}{N_{\text {total }}}=\frac{9}{115}=0.07826
\end{aligned}
$$ Noting that molar fraction is equal to pressure fraction, the partial pressure of water vapor is $$
P_{\mathrm{H} 2 \mathrm{O}}=y_{\mathrm{H} 2 \mathrm{O}} P_{\text {total }}=(0.07826)(101 \mathrm{kPa})=\mathbf{7 . 9 0 k P a}
$$ The temperature at which the condensation starts is the saturation temperature of water at this pressure. This is called the dew-point temperature. Then,
$$
T_{\text {cond }}=T_{\text {sat } @ 7.90 k \mathrm{~Pa}}=41.3^{\circ} \mathrm{C}
$$ Water vapor in the combustion gases will start to condense when the temperature of the combustion gases drop to $41.3^{\circ} \mathrm{C}$.
