Answer
According to Eq. $7-6,$
$$Y_{\mathrm{O}_{2}}=\frac{p \times \mathrm{O}_{2}}{K+(p \times \mathrm{O}_{2})}$$
When $$p \mathrm{O}_{2}=10 \text{ torr,}$$
$$Y_{\mathrm{O}_{2}}=\frac{10}{2.8+10}=0.78$$
When $$p \mathrm{O}_{2}=1 \text{ torr,}$$
$$Y_{\mathrm{O}_{2}}=\frac{1}{2.8+1}=0.26$$
Work Step by Step
According to Eq. $7-6,$
$$Y_{\mathrm{O}_{2}}=\frac{p \times \mathrm{O}_{2}}{K+(p \times \mathrm{O}_{2})}$$
When $$p \mathrm{O}_{2}=10 \text{ torr,}$$
$$Y_{\mathrm{O}_{2}}=\frac{10}{2.8+10}=0.78$$
When $$p \mathrm{O}_{2}=1 \text{ torr,}$$
$$Y_{\mathrm{O}_{2}}=\frac{1}{2.8+1}=0.26$$
