If 4.9g of nitrogen gas is placed in a 2 L container at 31 degrees, what is the pressure in the container?
In Kelvins, 31oC = 31 + 273 = 304 K. According to the ideal gas law,
PV=nRTPV=nRTPV=nRT
n(N2)=4.9 g×1 mol28.01 g=0.175 moln(N_2)=4.9\ g\times\frac{1\ mol}{28.01\ g}=0.175\ moln(N2)=4.9 g×28.01 g1 mol=0.175 mol
R=0.08206L⋅atmmol⋅KR=0.08206\frac{L\cdot{atm}}{mol\cdot{K}}R=0.08206mol⋅KL⋅atm (constant)
P=nRTV=0.175 mol×0.08206L⋅atmmol⋅K×304 K2 L=2.18 atmP=\frac{nRT}{V}=\frac{0.175\ mol\times0.08206\frac{L\cdot{atm}}{mol\cdot{K}}\times304\ K}{2\ L}=2.18\ atmP=VnRT=2 L0.175 mol×0.08206mol⋅KL⋅atm×304 K=2.18 atm
Answer: 2.18 atm
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