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A cylinder of compressed gas rolls off a boat and falls to the bottom of a lake. Eventually it rusts and the gas bubbles to the surface. A chemist collects a sample of the gas with the idea of trying to identify the gas. The wet gas collected occupies a volume of 285 mL at a pressure of 726 torr and temperature of 28.0oC. The vapor pressure of water at 28.0oC is 0.0372 atm.

1. Calculate the volume (L) that the gas occupies after it is dried (the water vapor removed) and stored at STP.

The mass of the dry gas is 214 mg. A fragment of the tank indicates that the gas is a monoatomic element.

2. Write out the full name of the gas.

Answer :

jufsanabriasa

Answer:

The volume that the gas occupies at STP is 237mL.

Gas is neon.

Explanation:

Using general gas law, the volume that the gas occupies after is dried at STP (273,15K; 1atm) is:

P1V1/T1 = P2V2/T2

The pressure of the wet gas is:

726torr×[tex]\frac{1atm}{760torr}[/tex] = 0,955atm

As vapor pressure of water is 0,0372atm, the pressure of the wet gas is:

0,955atm-0,0372atm = 0,9178atm.

Replacing:

0,9178atm×285mL/301,15K = 1atm×V2/273,15K

V2 = 237mL

To solve this problem it is necessary to obtain the moles of gas using gas law and, knowing the mass of the dry gas, it is possible to obtain the atomic mass of the element to know which one is.

Gas law is:

PV = nRT

n = PV/RT

Where P is pressure. The pressure of the wet gas is:

726torr×[tex]\frac{1atm}{760torr}[/tex] = 0,955atm

As vapor pressure of water is 0,0372atm, the pressure of the wet gas is:

0,955atm-0,0372atm = 0,9178atm.

V is volume: 0,285L.

R is gas constant: 0,082atmL/molK

And T is temperature (28°C + 273,15 = 301,15K)

Replacing:

n = 0,0106moles

As mass of dry tank is 214mg=0,214g. The atomic mass of the gas is:

0,214g/0,0106mol = 20,189 g/mol

Thus, gas is neon, because atomic mass of Ne is 20,18g/mol

I hope it helps!

batolisis
  1. The volume occupied by the dried gas  at STP is ; 237 mL
  2. The name of the gas is ; Neon

Using the Given data :

Applying gas law ;

[tex]\frac{P1V1}{T1 } = \frac{P2V2}{T2}[/tex] ----- ( 1 )

Pressure of the wet gas = 726 torr * ( 1 atm / 760 torr ) =  0.955 atm

Given that the vapor pressure of water = 0.0372atm,

∴ Actual pressure of the wet gas = (0.955 - 0.0372 ) = 0.9178atm.

Back to equation 1

[tex]\frac{P1V1}{T1 } = \frac{P2V2}{T2}[/tex]  

where : P₁ = 0.9178 atm,  V₁ = 285 mL , T₁ = 301.15 k,  P2 = 1 atm,

T₂ = 273.15K, V₂ = ?

Make V₂ subject of the equation then  Insert values into equation

∴ V₂ ( volume occupied by dry gas at STP ) = 237 mL

Next step : determine the atomic mass of the element using the mass of the dry gas and number of moles

Pv = nRT

n = Pv / RT -------- ( 2 )

where :  

P = pressure of gas ( wet ) = 0.9178 atm.

V = 0.285 L.

gas constant ( R ) = 0.082 atm L/mol.k

T = 301.15 K

Input values into equation ( 2 )

n = 0.0106 mole

Therefore the atomic mass of the gas = mass / n

                                                               = 0.214 g / 0.0106 mole

                                                                = 20,189 g/mol.

The gas with an atomic mass of 20.189 g/mol is Neon hence we can conclude that the gas is Neon. and the volume occupied by the dried gas  at STP is ; 237 mL

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