Respuesta :

Gay Lussac's law relates the pressure and temperature of a ideal gas in two different states at constant volume. In this case, they present us with a single state, they give us the grams of water vapor, so to find the volume we apply the ideal gas law. The ideal gas law has the following equation:

[tex]\begin{gathered} PV=nRT \\ V=\frac{nRT}{P} \end{gathered}[/tex]

Where,

V is the volume of the gas

n is the number of moles,

[tex]\begin{gathered} n=65gH_2O_v\times\frac{1molH_2O_v}{MolarMass,gH_2O_v} \\ n=65gH_2O_v\times\frac{1molH_2O_v}{18.01gH_2O_v}=3.6molH_2O_v \end{gathered}[/tex]

R is a constant, 0.08206atm.L/mol.K

T is the temperature, at STP the temperature is 273.15K

P is the pressure, at STP the pressure is 1 atm

We replace the known data and find the volume:

[tex]V=\frac{nRT}{P}[/tex][tex]V=\frac{3.6molH_2O\times0.08206\frac{atm.L}{mol.K}\times273.15K}{1atm}[/tex][tex]V=81L[/tex]

Answer: The volume of 65 grams of water at STP is 81 Liters

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