Question #31fed

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Answer 1 · 2014-08-09T17:06:23

The volume of dry hydrogen would have been 391.6 mL.

This problem involves Dalton's Law of Partial Pressures and Boyle's Law.

#P_"t"# = 740.0 Torr

At 18.0 °C, #P_"H₂O"# = 15.5 Torr.

#P_"t" = P_"H₂" + P_"H₂O"#

# P_"H₂" = P_"t" - P_"H₂O"# =(740.0 – 15.5) Torr = 724.5 Torr

Now we use Boyle's Law.

#P_1V_1 = P_2V_2#

#P_1# = 724.5 Torr; #V_1# = 400.0 mL
#P_2# = 740.0 Torr; #V_2# = ?

#V_2 = V_1 × P_1/P_2# = 400.0 mL × #(724.5"Torr")/(740.0"Torr")# = 391.6 mL

This makes sense. We increased the pressure by about 2 %, so the volume should decrease by about 2 % (8 mL).