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Worked Examples · Example 5.4

Q.A neon-dioxygen mixture contains 70.6 g dioxygen and 167.5 g neon. If pressure of the mixture of gases in the cylinder is 25 bar. What is the partial pressure of dioxygen and neon in the mixture ?

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Step 1 – Moles of each gas

nO2=70.6 g32 g mol−1=2.206 moln_{\text{O}_2} = \frac{70.6\ \text{g}}{32\ \text{g mol}^{-1}} = 2.206\ \text{mol}

nNe=167.5 g20.18 g mol−1=8.300 moln_{\text{Ne}} = \frac{167.5\ \text{g}}{20.18\ \text{g mol}^{-1}} = 8.300\ \text{mol}

Step 2 – Total moles and mole fractions

ntotal=2.206+8.300=10.506 moln_{\text{total}} = 2.206 + 8.300 = 10.506\ \text{mol}

xO2=2.20610.506=0.2100,xNe=8.30010.506=0.7900x_{\text{O}_2} = \frac{2.206}{10.506} = 0.2100, \qquad x_{\text{Ne}} = \frac{8.300}{10.506} = 0.7900

Step 3 – Apply Dalton's Law of Partial Pressures

pi=xi ptotalp_i = x_i \, p_{\text{total}}

pO2=0.2100×25 bar=5.25 barp_{\text{O}_2} = 0.2100 \times 25\ \text{bar} = 5.25\ \text{bar} …

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