What is the predicted density of nitrogen gas (N2) at 1.00atm and 298.15K?
hint: R=0.0821(L*atm)/(mol*K)
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
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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)
calculate the density of nitrogen gas molar mass n2 28.0g/mol r=0.0821
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