Question

Given the following sets of values for three of the gas variables, calculate the unknown quantity. a. \(P=782.4 \mathrm{~mm} \mathrm{Hg} ; V=? ; n=0.1021 \mathrm{~mol} ; T=26.2^{\circ} \mathrm{C}\) b. \(P=? \mathrm{~mm} \mathrm{Hg} ; V=27.5 \mathrm{~mL} ; n=0.007812 \mathrm{~mol} ; T=16.6^{\circ} \mathrm{C}\) c. \(P=1.045\) atm \(; V=45.2 \mathrm{~mL} ; n=0.002241 \mathrm{~mol} ; T=? \mathrm{C}\)

Short answer

In conclusion: a. The missing volume is \(V = 2.4682 \mathrm{~L}\) b. The missing pressure is \(P = 5522.5 \mathrm{~mm} \mathrm{Hg}\) c. The missing temperature is \(T = -2.36^{\circ} \mathrm{C}\)

Step-by-step solution

Before using the Ideal Gas Law, we should make sure all units are in their correct form. a. To convert °C to Kelvin: K = °C + 273.15 b. To convert mmHg to atm: 1 atm = 760 mmHg c. Convert mL to L: 1 L = 1000 mL The ideal gas constant for this problem is \(R = 0.0821 \dfrac{\mathrm{L} \cdot \mathrm{atm}}{\mathrm{K} \cdot \mathrm{mol}}\) #Step 2: Calculate the missing variable for each case#

Use the Ideal Gas Law to calculate the missing variable in each case. a. \(P = 782.4 \mathrm{~mm} \mathrm{Hg}\); \(V = ?\); \(n = 0.1021 \mathrm{~mol}\); \(T = 26.2^{\circ} \mathrm{C}\) - Convert \(P\) and \(T\): P = 782.4 mmHg * (1 atm / 760 mmHg) = 1.0295 atm; T = 26.2°C + 273.15 = 299.35 K - Use the Ideal Gas Law: \(V = \dfrac{nRT}{P} = \dfrac{0.1021 \mathrm{~mol} \times 0.0821 \mathrm{~L\cdot atm / (K \cdot mol)} \times 299.35 \mathrm{~K}}{1.0295 \mathrm{~atm}} = 2.4682 \mathrm{~L}\) b. \(P = ? \mathrm{~mm} \mathrm{Hg}\); \(V = 27.5 \mathrm{~mL}\); \(n = 0.007812 \mathrm{~mol}\); \(T = 16.6^{\circ} \mathrm{C}\) - Convert \(V\) and \(T\): V = 27.5 mL * (1 L / 1000 mL) = 0.0275 L; T = 16.6°C + 273.15 = 289.75 K - Use the Ideal Gas Law: \(P = \dfrac{nRT}{V} = \dfrac{0.007812 \mathrm{~mol} \times 0.0821 \mathrm{~L\cdot atm / (K \cdot mol)} \times 289.75 \mathrm{~K}}{0.0275 \mathrm{~L}} = 7.2717 \mathrm{~atm}\) - Convert \(P\) back to mmHg: P = 7.2717 atm * (760 mmHg / 1 atm) = 5522.5 mmHg c. \(P = 1.045 \mathrm{~atm}\); \(V = 45.2 \mathrm{~mL}\); \(n = 0.002241 \mathrm{~mol}\); \(T = ? \mathrm{C}\) - Convert \(V\): V = 45.2 mL * (1 L / 1000 mL) = 0.0452 L - Use the Ideal Gas Law for temperature in Kelvin: \(T = \dfrac{PV}{nR} = \dfrac{(1.045 \mathrm{~atm})(0.0452 \mathrm{~L})}{0.002241 \mathrm{~mol} \times 0.0821 \mathrm{~L\cdot atm / (K \cdot mol)}} = 270.79 \mathrm{~K}\) - Convert \(T\) back to °C: T = 270.79 K - 273.15 = -2.36°C In conclusion: a. The missing volume is \(V = 2.4682 \mathrm{~L}\) b. The missing pressure is \(P = 5522.5 \mathrm{~mm} \mathrm{Hg}\) c. The missing temperature is \(T = -2.36^{\circ} \mathrm{C}\)

Key concepts

Gas Laws Chemistry

Understanding gas laws in chemistry is crucial for explaining how gases behave under different conditions of temperature, pressure, and volume. These laws provide a foundational framework for interpreting the physical properties of gases and predicting their behavior. The gas laws are typically divided into several individual ones, such as Boyle’s Law, which shows the relationship between pressure and volume, and Charles's Law, which demonstrates how gases expand when heated. However, the Ideal Gas Law is particularly comprehensive, as it combines these simpler laws into one equation:
\( PV = nRT \).
Here, P stands for pressure, V indicates volume, n is the amount of substance in moles, R is the universal gas constant, and T is the temperature in Kelvin. This equation allows us to calculate any one of the variables if the others are known, making it a powerful tool in the study of chemistry.
Furthermore, it assumes that gas particles do not attract or repel each other and that they occupy no space themselves, which is mostly true for ideal gases but may need correction for real gases at high pressures or low temperatures.

Calculating Gas Variables

To calculate gas variables using the Ideal Gas Law, we need a clear understanding of the relationship between pressure, volume, temperature, and moles of gas. The equation makes it possible to solve for any one variable if the other three are known. For example, to find the volume when pressure, temperature, and moles of gas are given, we rearrange the formula to \( V = \frac{nRT}{P} \).
In the applied steps of an exercise, this formula allows us to determine the volume: \( V = 2.4682 \, L \) for a set of given conditions. The process requires not just arithmetic but also unit conversions to ensure that each variable matches the units of the gas constant. By methodically applying these calculations and paying close attention to units and significant figures, students can accurately determine the unknown variables in gas law problems. This mathematical approach is an essential skill in many real-world applications involving gases, such as in chemistry labs, industrial processes, and environmental science.

Converting Units in Chemistry

Converting units is an essential skill in chemistry, particularly when working with gas laws. It ensures that all variables are in the correct form and compatible with the gas constant used in equations. For instance, the Ideal Gas Law requires that pressure is in atmospheres (atm), volume in liters (L), temperature in Kelvin (K), and the amount of gas in moles (mol) when working with the most common gas constant \( R = 0.0821 \, \dfrac{L \cdot atm}{K \cdot mol} \).

• To convert Celsius to Kelvin, add 273.15 to the Celsius temperature.
• For pressure, convert mmHg to atm by dividing by 760, because 1 atm is equivalent to 760 mmHg.
• Volume conversions often involve changing milliliters (mL) to liters (L), which is done through division by 1000 since 1 L is 1000 mL.

In our exercise examples, such conversions were necessary to work with the Ideal Gas Law. This not only ensures accuracy but also teaches the significance of dimensional analysis – a technique widely used in chemistry to convert between units. Mastering unit conversions streamlines problem-solving and reinforces a thorough comprehension of chemistry concepts.