Question

At \(1450 . \mathrm{mmHg}\) and \(286 \mathrm{~K},\) a skin diver exhales a \(208-\mathrm{mL}\) bubble of air that is \(77 \% \mathrm{~N}_{2}, 17 \% \mathrm{O}_{2},\) and \(6.0 \% \mathrm{CO}_{2}\) by volume. (a) What would the volume (in \(\mathrm{mL}\) ) of the bubble be if it were exhaled at the surface at 1 atm and \(298 \mathrm{~K} ?\) (b) How many moles of \(\mathrm{N}_{2}\) are in the bubble?

Short answer

The final volume of the bubble at the surface is approximately 412.74 mL. The number of moles of \( \mathrm{N}_2 \) in the bubble is about 0.0130 moles.

Step-by-step solution

Convert Units for Pressure

Convert the initial pressure from mmHg to atm. Use the conversion factor: 1 atm = 760 mmHg.So, the initial pressure in atm is: \[ P_1 = \frac{1450}{760} \approx 1.91 \text{ atm} \]

Apply the Combined Gas Law

Use the Combined Gas Law to find the final volume of the bubble at the surface. The Combined Gas Law is given by: \[ \frac{P_1 V_1}{T_1} = \frac{P_2 V_2}{T_2} \]Substitute the known values:\[ P_1 = 1.91 \text{ atm}, V_1 = 208 \text{ mL}, T_1 = 286 \text{ K}, P_2 = 1 \text{ atm}, T_2 = 298 \text{ K} \]Solve for the final volume \(V_2\):\[ V_2 = \frac{P_1 V_1 T_2}{P_2 T_1} = \frac{1.91 \times 208 \times 298}{1 \times 286} \approx 412.74 \text{ mL} \]

Calculate the Moles of \( \mathrm{N}_2 \)

To find the moles of \(\mathrm{N}_2\), use the ideal gas law equation: \[ PV = nRT \]Rearrange the equation to solve for \(n\):\[ n = \frac{PV}{RT} \]First, calculate the partial pressure of \(\mathrm{N}_2\) in the bubble when exhaled:\[ P_{\mathrm{N}_2} = 0.77 \times P_1 = 0.77 \times 1.91 \text{ atm} \approx 1.47 \text{ atm} \]Then substitute the known values for \(P\), \(V\), \(R\), and \(T\):\[ P = 1.47 \text{ atm}, V = 0.208 \text{ L} (since 208 \text{ mL} = 0.208 \text{ L}), R = 0.0821 \text{ L·atm·K}^{-1}\text{·mol}^{-1}, T = 286 \text{ K} \]\[ n = \frac{1.47 \times 0.208}{0.0821 \times 286} \approx 0.0130 \text{ moles} \]

Key concepts

Gas laws in chemistry

Gas laws are essential in chemistry because they describe how gases behave under different conditions. When working with gases, it is crucial to understand how pressure, volume, and temperature interact. Three primary laws simplify these interactions: Boyle's Law, Charles's Law, and Avogadro's Law. Boyle's Law states that the volume of a gas is inversely proportional to its pressure at constant temperature. Charles's Law indicates that the volume of a gas is directly proportional to its temperature at constant pressure. Avogadro's Law tells us that the volume of a gas is proportional to the number of moles present at constant temperature and pressure. These laws combined form the Combined Gas Law, which is a useful tool for solving various gas-related problems in chemistry.

Ideal gas law

The Ideal Gas Law is a pivotal equation in chemistry that links four essential properties of gases: pressure (P), volume (V), temperature (T), and the number of moles (n). The equation is written as: \[ PV = nRT \]Here, \(R\) is the gas constant with a value of \(0.0821 \, \text{L·atm·K}^{-1}·\text{mol}^{-1}\). This law assumes that gases are ideal, meaning they have no intermolecular forces and occupy no volume. While real gases deviate slightly from ideal behavior, especially at high pressures and low temperatures, the Ideal Gas Law provides an excellent approximation for many practical purposes. In our exercise, we used the Ideal Gas Law to find the number of moles of \( \text{N}_2 \) in the exhaled bubble by rearranging the formula to solve for \( n \):\[ n = \frac{PV}{RT} \]

Partial pressure calculation

Partial pressure is the pressure exerted by a single gas component in a mixture of gases. To find the partial pressure, you multiply the total pressure of the mixture by the fraction of that gas in the mixture. In our example, the exhaled air bubble contains \( \text{N}_2 \), \( \text{O}_2 \), and \( \text{CO}_2 \) in specific ratios. To find the partial pressure of \( \text{N}_2 \), we calculated: \[ P_{ \text{N}_2} = 0.77 \times 1.91 \, \text{ atm} \approx 1.47 \, \text{ atm} \] This concept is critical in various applications, like calculating the amount of each gas component in a scuba diver's tank or understanding how gases exchange in the lungs.

Unit conversion in chemistry

Unit conversion is essential in chemistry because it allows measurements to be standardized, making calculations straightforward and comparable. For pressure, common units include atmospheres (atm), millimeters of mercury (mmHg), and pascals (Pa). To convert from mmHg to atm, we use the ratio \(1 \, \text{atm} = 760 \, \text{mmHg} \). For volume, we often convert between milliliters (mL) and liters (L), where \( 1 \, \text{L} = 1000 \, \text{mL} \). Accurate unit conversions ensure the correctness of calculations and the applicability of equations. For example, in our problem, we converted 1450 mmHg to atm by calculating: \[ P_1 = \frac{1450}{760} \approx 1.91 \, \text{atm} \] and converted 208 mL to L: \[ 208 \, \text{mL} = 0.208 \, \text{L} \].