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zloy xaker [14]
3 years ago
9

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.
Chemistry
1 answer:
dimulka [17.4K]3 years ago
5 0

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.

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1.89 x 10⁵ J = 1.89 x 10⁵ /1000 = 189.2 KJ

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