Chapter 5, Problem 1SAQ

Interpretation Introduction

To determine: The pressure (in atm) when volume of the sample is decreased to $225mL$.

Answer to Problem 1SAQ

Solution: $\left(b\right)1.60atm$.

Explanation of Solution

Given:

Initial pressure, P₁ $=547mmHg$

Initial volume, V₁ $=0.500L$

Final volume, V₂ $=225mL=0.225L\left(\because 1L=1000mL\right)$

According to Boyle’s law,

$P_1{V}_1=P_2{V}_2$

Substitute all the given values in the above equation to get

$547\text{mmHg}\times 0.500\text{L}=P_2\times 0.225\text{L}$

$\begin{array}{c}\Rightarrow P_{\text{2}}\text{=}\frac{\text{547}\text{mmHg}\text{×}\text{0}\text{.500}\cancel{\text{L}}}{\text{0}\text{.225}\cancel{\text{L}}}\\ P_{\text{2}}\text{=}\frac{\text{273}\text{.5}}{\text{0}\text{.225}}\text{mmHg}\\ P_{\text{2}}\text{=}\text{1215}\text{.55}\text{mmHg}\\ \end{array}$

To convert $1215.5mmHg$ into atmospheres, use the following relation:

$\therefore 760mmHg=1atm$

$or1mmHg=\frac{1}{760}atm$

$\Rightarrow 1215.5mmHg=\frac{1215.5}{760}atm\approx 1.60atm$

Therefore, the final pressure is $1.60atm$.

Conclusion

The pressure at which the volume of the sample is decreased to $225mL$ is $1.60atm$.