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.

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Solution

$\begin{array}{l} Here, \\ V = 0 {.166 m}^{3} \\ T = 300 K \\ {Mass of O}_{2} = 1.60 g \\ M_{O} = 32 g \\ n_{O} = \frac{1.60}{32} = 0.05 \\ {Mass of N}_{2} = 2.80 g \\ M_{N} = 28 g \\ n_{N} = \frac{2.80}{28} = 0.1 \\ {Partial pressure of O}_{2} is given by \\ P_{O} = \frac{n_{O} R T}{V} = \frac{0.05 \times 8.3 \times 300}{0.166} = 750 \\ {Partial pressure of N}_{2} is given by \\ P_{N} = \frac{n_{N} R T}{V} = \frac{0.1 \times 8.3 \times 300}{0.166} = 1500 \\ Total pressure is sum of the partial pressures. \\ \Rightarrow P = P_{N} + P_{O} = 750 + 1500 = 2250 P a \end{array}$

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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.

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.