Question

For each of the following sets of pressure/volume data, calculate the missing quantity. Assume that the temperature and the amount of gas remain constant. a. $V=19.3\mathrm{L}$ at $102.1\mathrm{kPa};V=10.0\mathrm{L}$ at $?\mathrm{kPa}$ b. $V=25.7\mathrm{mL}$ at 755 torr; $V=?$ at $761\mathrm{mm}\mathrm{Hg}$ c. $V=51.2\mathrm{L}$ at $1.05\mathrm{atm};V=?$ at $112.2\mathrm{kPa}$

Short answer

a. The final pressure is approximately $197.0\mathrm{kPa}$. b. The final volume is approximately $25.5\mathrm{mL}$. c. The final volume is approximately $48.6\mathrm{L}$.

Step-by-step solution

Problem a:

We are given the initial volume $V_1=19.3\mathrm{L}$ and the initial pressure $P_1=102.1\mathrm{kPa}$. We are also given the final volume $V_2=10.0\mathrm{L}$ and we need to find the final pressure $P_2$. Using Boyle's Law: $P_1{V}_1=P_2{V}_2$ Substitute the given values into the equation and solve for $P_2$: $102.1\mathrm{kPa}\times 19.3\mathrm{L}=P_2\times 10.0\mathrm{L}$ $P_2=\frac{102.1\mathrm{kPa}\times 19.3\mathrm{L}}{10.0\mathrm{L}}$ $P_2=197.050\mathrm{kPa}$ The final pressure is approximately $197.0\mathrm{kPa}$.

Problem b:

We are given the initial volume $V_1=25.7\mathrm{mL}$ and the initial pressure $P_1=755\mathrm{torr}$. We are also given the final pressure $P_2=761\mathrm{mm}\mathrm{Hg}$ and we need to find the final volume $V_2$. Since 1 torr is equal to 1 mmHg, we can directly use the given values in Boyle's Law: $P_1{V}_1=P_2{V}_2$ Substitute the given values into the equation and solve for $V_2$: $755\mathrm{torr}\times 25.7\mathrm{mL}=761\mathrm{torr}\times V_2$ $V_2=\frac{755\mathrm{torr}\times 25.7\mathrm{mL}}{761\mathrm{torr}}$ $V_2=25.4928\mathrm{mL}$ The final volume is approximately $25.5\mathrm{mL}$.

Problem c:

We are given the initial volume $V_1=51.2\mathrm{L}$ and the initial pressure $P_1=1.05\mathrm{atm}$. We are also given the final pressure $P_2=112.2\mathrm{kPa}$ and we need to find the final volume $V_2$. First, we need to convert the given pressure values into the same unit, such as kPa. 1 atm = 101.325 kPa, therefore: $P_1=1.05\mathrm{atm}\times 101.325\mathrm{kPa}/\mathrm{atm}=106.39125\mathrm{kPa}$. Now use Boyle's Law: $P_1{V}_1=P_2{V}_2$ Substitute the given and converted values into the equation and solve for $V_2$: $106.39125\mathrm{kPa}\times 51.2\mathrm{L}=112.2\mathrm{kPa}\times V_2$ $V_2=\frac{106.39125\mathrm{kPa}\times 51.2\mathrm{L}}{112.2\mathrm{kPa}}$ $V_2=48.5951\mathrm{L}$ The final volume is approximately $48.6\mathrm{L}$.

Key concepts

Understanding Gas Laws

To fully understand how gases behave under various conditions, scientists have developed a series of mathematical models known as the 'gas laws.' These laws explain the relationships between volume, pressure, temperature, and number of moles in a gas, assuming ideal behavior. Boyle's Law, one of these fundamental principles, is particularly focused on the pressure-volume relationship observed in a closed system when the temperature is held constant.

Gases consist of particles moving in random motion, colliding with each other and with the walls of their container. The force exerted by gas particles during these collisions manifests as pressure. Boyle's Law states that the pressure exerted by a gas held at a constant temperature varies inversely with the volume it occupies. This relationship is crucial for many scientific calculations and real-world applications, such as breathing mechanisms, syringes, and even scuba diving.

Pressure-Volume Relationship

Boyle's Law describes the pressure-volume relationship in gases and can be summarized by the equation
$P_1{V}_1=P_2{V}_2$.

This equation indicates that the product of the initial pressure ($P_1$) and volume ($V_1$) of a gas is equal to the product of its final pressure ($P_2$) and volume ($V_2$), assuming the temperature remains constant. This inverse relationship means that as the volume of the gas decreases, its pressure increases, and vice versa. Understanding this concept is essential, for example, when you're applying it to predict how changes in pressure will affect the volume of a gas and how to accurately manipulate these variables during laboratory work.

Units Conversion in Chemistry

Working with gases often requires converting between different units of pressure. Chemistry uses units such as atmospheres (atm), kilopascals (kPa), and millimeters of mercury (mmHg or torr) to measure pressure.

It is essential to ensure that all pressures are in the same unit before using them in calculations. For instance, to convert from atmospheres to kilopascals, you would use the fact that 1 atm equals 101.325 kPa. Precise units conversion is vital for the accuracy of Boyle's Law calculations and other scientific computations. Knowing how to convert between these units allows for a more profound understanding of the problems being solved and makes it possible to compare different data sets with ease.