A 497.9-gallon steel storage tank contains 91.7kg of methane. If the temperature is 23.7˚C what is the pressure inside the tank?

74.08 atm

Explanation:

We'll begin by converting 91.7 kg to grams (g). This can be obtained as follow:

1 kg = 1000 g

Therefore,

91.7 kg = 91.7 kg × 1000 g / 1 kg

91.7 kg = 91700 g

Next, we shall determine the number of mole present in 91700 g of methane, CH₄. This can be obtained as follow:

Mass of CH₄ = 91700 g

Molar mass of CH₄ = 12 + (4"1)

= 12 + 4

= 16 g/mol

Mole of CH₄ =?

Mole = mass /Molar mass

Mole of CH₄ = 91700 / 16

Mole of CH₄ = 5731.25 moles

Next, we shall convert 497.9 gallon to litres (L). This can be obtained as follow:

1 gallon = 3.785 L

Therefore,

497.9 gallon = 497.9 × 3.785

497.9 gallon = 1884.55 L

Next, we shall convert 23.7 ˚C to Kelvin temperature. This can be obtained as follow:

T(K) = T(˚C) + 273

T(˚C) = 23.7 ˚C

T(K) = 23.7 °C + 273

T(K) = 296.7 K

Finally, we shall determine the pressure in the tank as follow:

Number of mole (n) = 5731.25

Volume = 1884.55 L

Temperature (T) = 296.7 K

Gas constant (R) = 0.0821 atm.L/Kmol

Pressure (P) =?

PV = nRT

P × 1884.55 = 5731.25 × 0.0821 × 296.7

P × 1884.55 = 139607.92

Divide both side by 1884.55

P = 139607.92 / 1884.55

P = 74.08 atm

Therefore the pressure in the tank is 74.08 atm

120,630,890

step-by-step explanation:

110+10=120,620+10=630,880+10=890

b is the correct answer

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