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Wet helium gas is placed into a balloon at 24.4 degrees Celsius and a pressure of 765.3 mm Hg. What volume (in L) does the dry gas occupy if the water vapor pressure is 24.3 torr and the mass of dried helium gas in the balloon is 0.498 g

Answer :

Answer:

Hence the Volume of Gas = 3.04 L.

Explanation:  

pressure of dry gas = 765.3 - 24.3 = 741 mmhg  

Temperature of gas = 24.4+273.15 = 297.55 k  

No of mol of gas = 0.498/4 = 0.1245 mol  

R = gas constant = 0.0821 l.atm.k-1.mol-1  

From ideal-gas equation

PV = nRT  

(741/760) x v = 0.1245 x 0.0821 x 297.55  

V = Volume of Gas = 3.04 L

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