Question

Determine the final pressure when (a) \(7.50 \mathrm{~mL}\) of krypton at \(2.00 \times 10^{5} \mathrm{kPa}\) is transferred to a vessel of volume \(1.0 \mathrm{~L}\); (b) \(54.2 \mathrm{~cm}^{3}\) of \(\mathrm{O}_{2}\) at 643 Torr is compressed to \(7.8 \mathrm{~cm}^{3}\). Assume constant temperature.

Short answer

The final pressure for (a) krypton is 1500 kPa and for (b) oxygen is 596.7 kPa.

Step-by-step solution

- Understand the Gas Law to Use

For each part of this exercise, we need to use the combined gas law, which relates pressure, volume, and temperature of a gas that remains constant in amount and at constant temperature. The combined gas law is given by: \( \frac{P_1V_1}{P_2V_2} = \frac{T_1}{T_2} \). Since temperature is constant, the equation reduces to Boyle's law, \( P_1V_1 = P_2V_2 \), which we will use to find the final pressures.

- Calculate Final Pressure for Krypton

Given initial volume \(V_1 = 7.50 \mathrm{~mL} = 0.0075 \mathrm{~L} \) and initial pressure \(P_1 = 2.00 \times 10^5 \mathrm{kPa}\), and final volume \(V_2 = 1.0 \mathrm{~L}\), we apply Boyle's law to find \(P_2\). Rearranging the formula to solve for \(P_2\) yields: \[P_2 = \frac{P_1V_1}{V_2}\]. Substitute the known values to find \(P_2\).

- Calculate Final Pressure for Krypton (Cont'd)

\[P_2 = \frac{(2.00 \times 10^5 \mathrm{kPa})(0.0075 \mathrm{~L})}{1.0 \mathrm{~L}}\] Performing the calculation gives us \[P_2 = 1500 \mathrm{kPa}\].

- Convert Oxygen Volume to Liters

Before applying Boyle's Law to oxygen, convert its volume from \(cm^3\) to \(L\) as 1 L = 1000 cm\textsuperscript{3}. So, \(V_1 = 54.2 \mathrm{~cm}^{3} = 0.0542 \mathrm{~L}\) and \(V_2 = 7.8 \mathrm{~cm}^{3} = 0.0078 \mathrm{~L}\).

- Convert Oxygen Initial Pressure to kPa

The initial pressure is given in Torr. To use the same units, convert to kPa using the conversion 1 atm = 760 Torr and 1 atm = 101.325 kPa. Thus, \[P_1 = 643 \mathrm{Torr} = \frac{643}{760} \mathrm{atm} = \frac{643}{760} \times 101.325 \mathrm{kPa}\].

- Calculate Final Pressure for Oxygen

With the initial pressure in kPa and volumes in litres, apply Boyle's law. First, calculate \(P_1\) in kPa, then use \(P_1V_1 = P_2V_2\) to solve for \(P_2\).

- Calculate Final Pressure for Oxygen (Cont'd)

First calculate \(P_1\) in kPa: \[P_1 = (643 \times 101.325) / 760 = 86.06842105 \mathrm{kPa}\] Now solve for \(P_2\): \[P_2 = \frac{P_1V_1}{V_2}\] \[P_2 = \frac{(86.06842105 \mathrm{kPa})(0.0542)}{0.0078}\] Perform the calculation to achieve \[P_2 = 596.7 \mathrm{kPa}\].

Key concepts

Combined Gas Law

Understanding the Combined Gas Law is critical when exploring the behavior of gases under different conditions. This law is a blend of three fundamental gas laws: Boyle's Law (pressure-volume relationship), Charles' Law (volume-temperature relationship), and Gay-Lussac's Law (pressure-temperature relationship).

The formula for the Combined Gas Law is \( P_1V_1/T_1 = P_2V_2/T_2 \) where \( P \) stands for pressure, \( V \) for volume, and \( T \) for temperature, with subscripts 1 and 2 representing the initial and final states of the gas, respectively. When temperature is constant, as in our exercise, the Combined Gas Law simplifies to Boyle's Law, focusing solely on the inverse relationship between pressure and volume.

Pressure-Volume Relationship

The relationship between pressure and volume of a gas at a constant temperature is explained by Boyle's Law, which is a core concept in the study of gases. It states that for a given mass of an enclosed gas, pressure is inversely proportional to volume. Mathematically, Boyle's Law is expressed as \( P_1V_1 = P_2V_2 \).

This law signifies that if we increase the volume of the gas container, the pressure exerted by the gas decreases, and conversely, reducing the volume increases pressure. It's important to note that Boyle's Law assumes that the temperature and amount of gas remain unchanged throughout the process.

Gas Constant Temperature

In gas law calculations, constant temperature is often a key condition, typically represented by 'isothermal' processes. Maintaining gas at a constant temperature means that any changes made to the volume or pressure will not affect the overall temperature of the system. In scenarios where temperature could change, the ideal gas constant would come into play, which is a part of the Ideal Gas Law. However, for exercises dealing only with Boyle's Law, the temperature factor is removed, allowing for straightforward solutions focusing on just pressure and volume.

Units Conversion

Working with gases, we often need to convert units to maintain consistency and accuracy in calculations. In the case of pressure, common units include pascals (Pa), kilopascals (kPa), atmospheres (atm), and torr or mmHg. For volume, units such as liters (L), milliliters (mL), and cubic centimeters (cm\(^3\)) are frequently used.

For instance, to convert volume from milliliters to liters, we use the conversion factor \( 1 \text{L} = 1000 \text{mL} \). Similarly, pressure conversions may involve converting torr to kPa, which is facilitated by the relationship \( 1 \text{atm} = 760 \text{torr} \) and \( 1 \text{atm} = 101.325 \text{kPa} \). These conversions are crucial for ensuring that all terms in gas law equations are in compatible units.