(2.07kPa-y_(i))/(2.07kPa-1.93kPa)=(18.0^circ C-17.6^circ C)/(18.0^circ )C-17.0^(circ C)

Here's how to solve for y_i:<br /><br />1. **Simplify the fractions:**<br /><br /> $\frac {2.07kPa-y_{i}}{0.14kPa}=\frac {0.4^{\circ }C}{1.0^{\circ }C}$<br /><br />2. **Further simplify the right side:**<br /><br /> $\frac {2.07kPa-y_{i}}{0.14kPa}=0.4$<br /><br />3. **Multiply both sides by 0.14 kPa:**<br /><br /> $2.07kPa - y_{i} = 0.4 * 0.14kPa$<br /><br />4. **Calculate the right side:**<br /><br /> $2.07kPa - y_{i} = 0.056kPa$<br /><br />5. **Isolate y_i:**<br /><br /> $y_{i} = 2.07kPa - 0.056kPa$<br /><br />6. **Calculate y_i:**<br /><br /> $y_{i} = 2.014kPa$<br /><br />Therefore, $y_{i} = 2.014 kPa$<br />