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The Ideal Gas Law is one of the most fundamental equations in chemistry, used to relate the pressure, volume, temperature, and number of moles of a gas. It is expressed as:
\[ PV = nRT \]
Where:
- \( P \) is the pressure of the gas (in atmospheres when using the universal gas constant R),
- \( V \) is the volume of the gas (in liters),
- \( n \) is the number of moles,
- \( R \) is the universal gas constant (0.0821 L atm K鈦宦� mol鈦宦�),
- \( T \) is the temperature (in Kelvin).
This law helps us to determine either the unknown property of a gas or find out how the gas behaves under different conditions. In the exercise above, we used the Ideal Gas Law to determine the number of moles of \( \mathrm{CO}_{2} \) and \( \mathrm{H}_{2} \mathrm{O} \) exhaled. By rearranging the formula to solve for moles \( n \), we calculated:
\[ n = \frac{PV}{RT} \] We used the given partial pressures, volume, temperature, and the gas constant to find \( n \).

In a mixture of gases, each gas exerts its own pressure as if it were alone in the container. This individual pressure is known as the partial pressure. Dalton's Law of Partial Pressures states that the total pressure of a mixture of gases is equal to the sum of the partial pressures of the individual component gases.
For instance, in the exercise from above, the partial pressures of \( \mathrm{CO}_{2} \) and \( \mathrm{H}_{2} \mathrm{O} \) in the exhaled air are both given as 30.0 torr.
To convert these pressures to atmospheres (since the universal gas constant \( R \) in the Ideal Gas Law uses atm), we converted as follows:
\[ \text{Pressure} (\text{atm}) = \text{Pressure} (\text{torr}) \times \frac{1 \text{ atm}}{760 \text{ torr}} \] By converting these pressures and using them in the Ideal Gas Law, we accurately calculated the number of moles of each gas.

Glucose metabolism is the process where glucose is converted into energy, with the chemical equation:
\[ \mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}(s) + 6 \mathrm{O}_{2}(g) \rightarrow 6 \mathrm{CO}_{2}(g) + 6 \mathrm{H}_{2} \mathrm{O}(g) \]
This reaction means that one mole of glucose produces six moles of carbon dioxide and six moles of water.
In the exercise, the total mass of \( \mathrm{CO}_{2} \) and \( \mathrm{H}_{2} \mathrm{O} \) exhaled comes from the metabolism of glucose. The calculation of body mass lost over 8 hours of sleep is based on these products of metabolism.
Understanding glucose metabolism helps us comprehend how energy production and gas exchange work together, impacting body mass due to respiratory processes during activities such as sleeping.

The molar mass of a substance is the mass of one mole of its entities (atoms, molecules, etc.) and is usually expressed in grams per mole (g/mol). It is essential for converting between the mass of a substance and the number of moles.
For example, the molar mass of carbon dioxide \( \mathrm{CO}_{2} \) and water \( \mathrm{H}_{2} \mathrm{O} \) was crucial in the exercise.
- \( \mathrm{CO}_{2} \) has a molar mass of 44.01 g/mol, calculated as:
\[ 12.01 \text{(C)} + 2 \times 16.00 \text{(O)} = 44.01 \text{ g/mol} \]
- \( \mathrm{H}_{2} \mathrm{O} \) has a molar mass of 18.02 g/mol, calculated as:
\[ 2 \times 1.01 \text{(H)} + 16.00 \text{(O)} = 18.02 \text{ g/mol} \]
Using these molar masses, we converted moles to grams, providing the final mass values needed for the calculation of how much body mass was lost over 8 hours of sleep.