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Question
A vessel contains 1.60 g of oxygen and 2.80 g of nitrogen. The temperature is maintained at 300 K and the volume of the vessel is 0.166 m3. Find the pressure of the mixture.
Use R = 8.3 J K-1 mol-1
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Solution
Here ,
V = 0 .166 m3
T = 300 K
Mass of O2 = 1.60 g
MO = 32 g
nO = \[\frac{1.60}{32} = 0.05 \]
Mass of N2 = 2.80 g
\[\ M_N = 28 g \]
\[\ n_N = \frac{2.80}{28} = 0.1 \]
Partial pressure of O2 is given by
\[\ P_O = \frac{n_O RT}{V} = \frac{0.05 \times 8.3 \times 300}{0.166} = 750 \]
Partial pressure of N2 is given by
\[\ P_N = \frac{n_N RT}{V} = \frac{0.1 \times 8.3 \times 300}{0.166} = 1500 \]
Total pressure is sum of the partial pressures.
⇒ P = PN + PO = 750 + 1500 = 2250 Pa
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