If 200 mL of N $_{2}$ at 25 $^{\circ} C$ and a pressure of 250 mm are mixed with 350 mL of O $_{2}$ at 25 $^{\circ} C$ and a pressure of 300 mm so that, the volume of resulting mixture is 300 mL, what would be the final pressure of the mixture at 25 $^{\circ} C$ ?

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Solution

In the case of nitrogen, volume increases, its pressure must decrease. Let the new pressure be $P_{N_{2}}$
$P_{N_{2}} = \frac{250 \times 200}{300} = 166.6 m m$ (Applying Boyle's law)
In the case of oxygen, volume decreases, its pressure must increase. Let the new pressure be $P_{O_{2}}$
$P_{O_{2}} = \frac{300 \times 350}{300} = 350 m m$ (Applying Boyle's law)
Total pressure $= P_{N_{2}} + P_{O_{2}} = (1666.6 + 350) = 516.6 m m$

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