Question

Using the Gay-Lussac LawWhat is the volume of 65 grams of water vapor at STP?

Answer 1

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:

`\begin{gathered} PV=nRT \\ V=(nRT)/(P) \end{gathered}`

Where,

V is the volume of the gas

n is the number of moles,

`\begin{gathered} n=65gH_2O_v*(1molH_2O_v)/(MolarMass,gH_2O_v) \\ n=65gH_2O_v*(1molH_2O_v)/(18.01gH_2O_v)=3.6molH_2O_v \end{gathered}`

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:

`V=(nRT)/(P)`
`V=(3.6molH_2O*0.08206(atm.L)/(mol.K)*273.15K)/(1atm)`
`V=81L`

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