Question

Given each of the following sets of values for an ideal gas, calculate the unknown quantity. a. \(P=782 \mathrm{mm} \mathrm{Hg} ; V=? ; n=0.210 \mathrm{mol} ; T=27^{\circ} \mathrm{C}\) b. \(P=? \mathrm{mm} \mathrm{Hg} ; V=644 \mathrm{mL} ; n=0.0921\mathrm{mol} ; T=\) \(303 \mathrm{K}\) c. \(P=745 \mathrm{mm} \mathrm{Hg} ; V=11.2 \mathrm{L} ; n=0.401 \mathrm{mol} ; T=\) ? K

Short answer

The calculated missing values for the ideal gas sets are as follows: a. V = 5.31 L b. P = 2674 mmHg c. T = 340 K

Step-by-step solution

Convert given values to proper units

First, we need to convert the given values into proper units as required by the Ideal Gas Law equation. The units to be used are: Pressure (P): atmospheres (atm) Volume (V): liters (L) Number of moles (n): moles (mol) Temperature (T): Kelvin (K) (The value of the gas constant R is often used as 0.0821 L atm / K mol, so we need to have all the units in the corresponding form) a. Convert P to atm and T to K P = 782 mmHg x (1 atm / 760 mmHg) = 1.029 atm T = 27°C + 273.15 = 300.15 K b. Convert V to L and P to the desired unit (mmHg) V = 644 mL x (1 L / 1000 mL) = 0.644 L c. Convert P to atm and V to L P = 745 mmHg x (1 atm / 760 mmHg) = 0.980 atm V = 11.2 L

Solve for the missing value using the Ideal Gas Law

Now that we have our values in the correct units, we can apply the Ideal Gas Law, PV = nRT, to each set of values to find the missing quantity. a. Solve for V V = nRT / P = (0.210 mol)(0.0821 L atm / K mol)(300.15 K) / 1.029 atm = 5.31 L b. Solve for P P = nRT / V = (0.0921 mol)(0.0821 L atm / K mol)(303 K) / 0.644 L = 3.52 atm Here, we need to convert P back to mmHg to match the desired units. P = 3.52 atm x (760 mmHg / 1 atm) = 2674 mmHg c. Solve for T T = PV / nR = (0.980 atm)(11.2 L) / (0.401 mol)(0.0821 L atm / K mol) = 340 K Now, let's summarize the calculated missing values: a. V = 5.31 L b. P = 2674 mmHg c. T = 340 K

Key concepts

Pressure Conversion

Pressure is an essential component of the Ideal Gas Law. When dealing with pressure, we often need to convert between different units to ensure consistency in calculations. The most common pressure units are atmospheres (atm), millimeters of mercury (mm Hg), and Pascals (Pa).

To convert from mm Hg to atm, use the relationship:

• 1 atm = 760 mm Hg.

Thus, to convert a pressure measurement in mm Hg to atm, you can use: \[P(\text{atm}) = P(\text{mm Hg}) \times \frac{1 \text{ atm}}{760 \text{ mm Hg}}.\]
For example, if a pressure is 782 mm Hg, it converts to 1.029 atm. This conversion ensures compatibility with the gas constant's units (R).

Understanding pressure conversion allows you to accurately apply the Ideal Gas Law for calculations in different contexts.

Volume Conversion

Volume is another crucial variable in the Ideal Gas Law, often measured in liters (L) or milliliters (mL). It's important to convert these units as needed to maintain consistency in calculations.

To convert volume from milliliters to liters, use the relationship:

• 1000 mL = 1 L.

Therefore, to convert a volume in mL to L, apply: \[V(\text{L}) = V(\text{mL}) \times \frac{1 \text{ L}}{1000 \text{ mL}}.\]
For instance, converting 644 mL gives 0.644 L. This aligns with the Ideal Gas Law's requirements, which often involve using volumes in liters.

Being proficient in volume conversion ensures proper functionality of calculations when using various gas laws.

Temperature Conversion

Temperature is another pivotal part of the Ideal Gas Law, which must be expressed in Kelvin (K) to coincide with the gas constant's units. In many problems, the given temperature will be in Celsius (°C), so conversion to Kelvin is essential.

To convert Celsius to Kelvin, use the formula:

• K = °C + 273.15.

For example, a temperature of 27°C becomes 300.15 K in Kelvin.

This conversion ensures that temperature values are compatible with the Ideal Gas Law equations, as calculations in Kelvin prevent negative temperature scenarios that are not physically meaningful in this context.
Emphasizing temperature conversion aids in preventing errors during computations involving gases.

Gas Constant R

The gas constant, denoted as \(R\), is a key factor in the Ideal Gas Law equation, allowing us to relate pressure, volume, temperature, and moles in a consistent manner.

Its value depends on the units used and is typically 0.0821 L atm / K mol, fitting the common units of liters, atmospheres, Kelvin, and moles.

• This means \(R\) is tailored to harmonize with the units of the measured quantities in the Ideal Gas Law.

Therefore, when solving problems, always ensure given data aligns with \(R\)'s units. If not, apply conversions for pressure, volume, or temperature as needed.

Understand that the careful alignment of units with \(R\) enables accurate calculations, making it a cornerstone of gas-related computations in chemistry.