Chapter 11, Problem 1SAQ

Interpretation Introduction

Interpretation: A gas sample has an initial pressure of $547\text{mmHg}$ and an initial volume of $\text{0}\text{.500 L}$ the pressure in atm should be determined when the volume of the sample is decreased by $\text{225}\text{mL}$ assume constant temperature and constant number of moles of gas.

Concept introduction: According to Boyle’s law, at constant temperature and constant number of moles of ideal gas the pressure of the gas is inversely proportional to its volume.

${\text{P}}_{\text{1}}{\text{V}}_{\text{1}}\text{=}{\text{P}}_{\text{2}}{\text{V}}_{\text{2}}$

Where P is pressure and V is the volume of the gas in state 1 and 2 respectively.

Correct answer:

Correct answer is (d) option (a), (b) and (c) incorrect answer

Explanation of Solution

Reason for correct option: According to Boyle’s law

${\text{P}}_{\text{1}}{\text{V}}_{\text{1}}\text{=}{\text{P}}_{\text{2}}{\text{V}}_{\text{2}}$

Here the initial pressure ${\text{(P}}_{\text{1}}\text{)}\text{is}\text{547mmHg,}$ initial volume ${\text{(V}}_{\text{1}}\text{)is}\text{0}\text{.500 L}$ and final volume is ${\text{(V}}_{\text{2}}\text{)}\text{is}\text{225}\text{mL}$ .

Converting 225 mL to L:

  $\begin{array}{l}{\text{V}}_{\text{2}}\text{=225 mL×}\frac{\text{1 L}}{\text{1000 mL}}\text{(1}\text{L}\text{=}\text{1000 mL)}\\ \text{=}\text{0}\text{.225 L}\end{array}$

To calculate the final pressure ${\text{(P}}_{\text{2}}\text{)}$ :

  $\begin{array}{l}{\text{P}}_{\text{1}}{\text{V}}_{\text{1}}{\text{ = P}}_{\text{2}}{\text{V}}_{\text{2}}\\ {\text{P}}_{\text{2}}\text{=}\frac{{\text{V}}_{\text{1}}}{{\text{V}}_{\text{2}}}{\text{P}}_{\text{1}}\\ \text{= }\frac{\text{0}\text{.500 L}}{\text{0}\text{.225 L}}\text{×547 mmHg}\\ \text{=1215}\text{.5 mmHg×}\frac{\text{1}\text{atm}}{\text{760 mmHg}}\text{(1}\text{atm=760mm}\text{Hg)}\\ \text{=1}\text{.60}\text{atm}\end{array}$

Thus the pressure will be $1.60\text{atm}$ and the correct answer will be option (d).