# Question #b7786

Aug 30, 2017

$600 c {m}^{3}$

#### Explanation:

for the Dalton law about partial pressure you must take off from the first conditions the vapour pressure of water (380 torr = 0,5 atm) and remains 1,5 atm. Now with the ideal gasses law (or with Boyle's law) you can change the conditions:
${P}_{1} \times {V}_{1} = {P}_{2} \times {V}_{2}$
Where STP are normally 1 atm.
${V}_{2} = {P}_{1} \times {V}_{1} / {P}_{2} = \frac{1 , 5 a t m \times 400 c {m}^{3}}{1 a t m} = 600 c {m}^{3}$

Aug 31, 2017

$787 , 6 c {m}^{3}$
${P}_{1} \times {V}_{1} = {P}_{2} \times {V}_{2}$
${V}_{2} = \frac{{V}_{1} \times {P}_{1}}{{P}_{2}} = \frac{400 c {m}^{3} \times 1 , 969 a t m}{1 a t m} = 787 , 6 c {m}^{3}$