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What is the predicted density of nitrogen gas (N2) at 1.00atm and 298.15K? hint: R=0.0821(L*atm)/(mol*K)

What is the predicted density of nitrogen gas (N2) at 1.00atm and 298.15K?

hint: R=0.0821(L*atm)/(mol*K)

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Answer #1

Answer:

From Ideal gas equation, PV=nRT

Where P=pressure, V=volume, R=gas constant, T=temperature,

n=number of moles =mass/molar mass.

PV=(mass/molar mass)RT

Mass/volume=(P x molar mass)/RT

Density=(P x molar mass)/RT (since density=mass/volume)

Given P= 1 atm, T=298.15 K, R=0.0821 L atm mol^-1 K^-1, molar mass of N2=28 g/mol.

Density=(1 atm x 28 g/mol)/(0.0821 L atm mol^-1 K^-1 x 298.15 K)

Density=1.144 g/L.

Therefore the predicted density of N2=1.144 g/L

Thanks and I hope you like it. Please let me know if you have any doubt.

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