# Question #72788

May 3, 2016

The aqueous tension is $0.1 a t m$

#### Explanation:

In order to find the aqueous tension (or vapour pressure) we can first figure out what pressure would the dry gas exert at ${10}^{\circ} C$ in a $0.1 L$ container?

Since the number of mole is constant, we can use the following expression to find the new pressure ${P}_{2}$:

$\frac{{P}_{1} {V}_{1}}{{T}_{1}} = \frac{{P}_{2} {V}_{2}}{{T}_{2}} \implies {P}_{2} = \frac{{P}_{1} {V}_{1}}{{T}_{1}} \times \frac{{T}_{2}}{{V}_{2}}$

$\implies {P}_{2} = \frac{1 a t m \times 0.080 \cancel{L}}{273 \cancel{K}} \times \frac{283 \cancel{K}}{0.1 \cancel{L}} = 0.8 a t m$ (1 significant figure since the volume 0.1L is one significant figure).

Therefore, the vapour pressure or aqueous tension would be found by:

$0.90 a t m - 0.8 a t m = 0.1 a t m$