0.38 g of liquid ethanol (C 2 H 5 OH) is placed in a 4.7-L bottle. The bottle is

Question

0.38 g of liquid ethanol (C2H5OH) is placed in a 4.7-L bottle. The bottle is sealed and placed in a refrigerator where the temperature is 0.0oC. How much liquid ethanol will be present in the bottle after liquid-vapor equilibrium is reached? The vapor pressure of ethanol is 10.0 torr at -2.3oC and 40.0 torr at 19oC.

0.38 g of liquid ethanol (C2H5 OH) is placed in a 4.7-L bottle. The bottle is sealed and placed in a refrigerator where the temperature is 0.0 C. How much liquid ethanol will be present in the bottle after liquid-vapor equilibrium is reached? The vapor pressure of ethanol is 10. torr at -2.3°C and 40. torr at 19 C

Explanation / Answer

Calculate the heat of vaporization from the Clausius-Clapeyron relation.
ln(P2/P1) = (Hvap/R)(1/T1 - 1/T2)
ln(40/10) = (Hvap/8.314 J/mol)(1/(-2.3 + 273K) - 1/(19 + 273K))
Hvap = 42772 J/mol-K
Calculate the vapor pressure of ethanol at 0C.
ln(P2/10) = (42772 J/mol-K / 8.314 J/mol)(1/(-2.3 + 273K) - 1/(0 + 273K))
P2 = 11.736 torr = 0.01544 atm
PV = nRT

0.01544 * 4.7 = n *0.0821*273

n = 3.23*10^-3 moles of vapor

moles of original ethanol = 0.38 / 46 =8.26*10^-3

moles of liquid state = 8.26*10^-3 - 3.23*10^-3 = 5.03*10^-3

liquid state ethanol = 0.23138 gms

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