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

Explanation:

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

The combined gas equation is:

$\frac{P_1V_1}{T_1}=\frac{P_2V_2}{T_2}$

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

$\frac{742.5mmHg\times 85.0ml}{293K}=\frac{760mmHg\times V_2}{273K}$

$V_2=77.4mL$

Volume of dry gas at STP is 77.4 mL.