Question

If the volume of wet gas collected over water is 85.0 mL at 20°C and 760 mm Hg , what is the volume of dry gas at STP conditions? (The vapor pressure of water at 20°C is 17.5 mm Hg.) Express your answer with the appropriate units.

Answer 1

Answer: 77.4 mL

Step-by-step explanation:

Combined gas law is the combination of Boyle's law, Charles's law and Gay-Lussac's law.

The combined gas equation is:

`(P_1V_1)/(T_1)=(P_2V_2)/(T_2)`

where,

`P_1` = initial pressure of dry gas = (760 - 17.5) mmHg= 742.5 mm Hg

`P_2` = final pressure of dry gas at STP = 760 mm Hg

`V_1` = initial volume of dry gas = 85.0 mL

`V_2` = final volume of dry gas at STP = ?

`T_1` = initial temperature of dry gas =
`20^oC=273+20=293K`

`T_2` = final temperature of dry gas at STP =
`0^oC=273+0=273K`

Now put all the given values in the above equation, we get the final volume of wet gas at STP

`(742.5mmHg* 85.0ml)/(293K)=(760mmHg* V_2)/(273K)`

`V_2=77.4mL`

Volume of dry gas at STP is 77.4 mL.