Assuming ideal behavior, how many moles of argon would you need to fill a 14.012.010.0 ft room? assume atmospheric pressure of 1.00 atm, a room temperature of 20.0 ∘c, and 28.2 l/ft3. express your answer with the appropriate units.

Question

Braedon Lindsey

Chemistry

15 May, 17:17

Assuming ideal behavior, how many moles of argon would you need to fill a 14.012.010.0 ft room? assume atmospheric pressure of 1.00 atm, a room temperature of 20.0 ∘c, and 28.2 l/ft3. express your answer with the appropriate units.

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Answers (1)

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Jayce Novak · Answer 1

We use the formula:

PV = nRT

First let us get the volume V:

volume = 14 ft * 12 ft * 10 ft = 1,680 ft^3

Convert this to m^3:

volume = 1680 ft^3 * (1 m / 3.28 ft) ^3 = 47.61 m^3

n = PV / RT

n = (1 atm) (47.61 m^3) / (293.15 K * 8.21x10^-5 m3 atm / mol K)

n = 1,978.13 mol

Assuming ideal behavior, how many moles of argon would you need to fill a 14.0*12.0*10.0 ft room? assume atmospheric pressure of 1.00 atm, a room temperature of 20.0 ∘c, and 28.2 l/ft3. express your answer with the appropriate units.

Assuming ideal behavior, how many moles of argon would you need to fill a 14.012.010.0 ft room? assume atmospheric pressure of 1.00 atm, a room temperature of 20.0 ∘c, and 28.2 l/ft3. express your answer with the appropriate units.