Question

Given each of the following sets of values for three of the gas variables, calculate the unknown quantity. a. \(P=7.74 \times 10^{3} \mathrm{Pa} ; V=12.2 \mathrm{mL} ; n=? \mathrm{mol} ; T=\) \(298 \mathrm{K}\) b. \(P=? \mathrm{mm} \mathrm{Hg} ; V=43.0 \mathrm{mL} ; n=0.421 \mathrm{mol} ; T=\) \(223 \mathrm{K}\) c. \(\bar{P}=455 \mathrm{mm}\) Hg; \(V=? \mathrm{mL} ; n=4.4 \times 10^{-2} \mathrm{mol}\) \(T=331^{\circ} \mathrm{C}\)

Short answer

The short answers for the given problems are: a. For the unknown quantity n, the calculated value is \(n ≈ 0.0040\,\text{mol}\). b. For the unknown quantity P, the calculated value is \(P ≈ 681.40\,\text{mmHg}\). c. For the unknown quantity V, the calculated value is \(V ≈ 0.298\,\text{L}\).

Step-by-step solution

Write down the known values and the unknown value

We have: P = 7.74 × 10^3 Pa V = 12.2 mL (we will convert this to liters) n = ? T = 298 K We need to find the value of n.

Convert units if necessary

Volume should be in liters. So, let's convert mL to L: \(12.2\,\text{mL} = 12.2 \times 10^{-3}\,\text{L}\)

Rearrange the Ideal Gas Law to solve for n

We have: \(PV = nRT\) Now rearrange it for n: \(n = \frac{PV}{RT}\)

Substitute known values and solve for n

Now substitute the known values and the value of R (8.314 J/(mol*K)): \(n = \frac{(7.74 \times 10^3\,\text{Pa})(12.2 \times 10^{-3}\,\text{L})}{(8.314\frac{\text{J}}{\text{mol}\cdot\text{K}})(298\,\text{K})}\) Now calculate n: \(n ≈ 0.0040\,\text{mol}\) #Problem 2: Find P#

Write down the known values and the unknown value

We have: P = ? mmHg V = 43.0 mL (we will convert this to liters) n = 0.421 mol T = 223 K We need to find the value of P.

Convert units if necessary

Volume should be in liters. So, let's convert mL to L: \(43.0\,\text{mL} = 43.0 \times 10^{-3}\,\text{L}\)

Rearrange the Ideal Gas Law to solve for P

We have: \(PV = nRT\) Now rearrange it for P: \(P = \frac{nRT}{V}\)

Substitute known values and solve for P

Now substitute the known values and the value of R (62.364 L*mmHg/(mol*K)): \(P = \frac{(0.421\,\text{mol})(62.364\frac{\text{L}\cdot\text{mmHg}}{\text{mol}\cdot\text{K}})(223\,\text{K})}{43.0 \times 10^{-3}\,\text{L}}\) Now calculate P: \(P ≈ 681.40\,\text{mmHg}\) #Problem 3: Find V#

Write down the known values and the unknown value

We have: P = 455 mmHg V = ? L n = 4.4 × 10^-2 mol T = 331°C (we will convert this to Kelvin) We need to find the value of V.

Convert units if necessary

Temperature should be in Kelvin. So, let's convert °C to K: \(331\,^{\circ}\text{C} = 331 + 273.15 = 604.15\,\text{K}\)

Rearrange the Ideal Gas Law to solve for V

We have: \(PV = nRT\) Now rearrange it for V: \(V = \frac{nRT}{P}\)

Substitute known values and solve for V

Now substitute the known values and the value of R (62.364 L*mmHg/(mol*K)): \(V = \frac{(4.4 \times 10^{-2}\,\text{mol})(62.364\frac{\text{L}\cdot\text{mmHg}}{\text{mol}\cdot\text{K}})(604.15\,\text{K})}{455\,\text{mmHg}}\) Now calculate V: \(V ≈ 0.298\,\text{L}\)

Key concepts

Gas Variables

Gas variables are key elements in understanding how gases behave and interact with their environment. These variables include:

• Pressure (P): This measures the force that the gas exerts on the walls of its container. Common units are Pascals (Pa) and millimeters of mercury (mmHg).
• Volume (V): This represents the space occupied by the gas. It is typically measured in liters (L) or milliliters (mL).
• Moles (n): This is the amount of substance measured in mole (mol), referring to the quantity of gas present.
• Temperature (T): It indicates how hot or cold the gas is, normally recorded in Kelvin (K).

Understanding these variables is crucial as they all interrelate according to the Ideal Gas Law. This law describes the state of an ideal gas through the equation: \[PV = nRT\]Where \(R\) is the ideal gas constant, different values of which are used depending on the units of pressure.

Unit Conversion

In scientific measurements, unit conversion is essential for consistency and accuracy. Let's explore this further:

• Volume Conversion: Often, volume is given in milliliters (mL) but needs to be converted to liters (L) as the standard metric unit of volume. To convert, use the relation: \[1 \, ext{mL} = 1 imes 10^{-3} \, ext{L}\]For example, \(12.2 \, ext{mL}\) would convert to \(12.2 imes 10^{-3} \, ext{L}\).
• Temperature Conversion: Since gases depend on temperature, converting Celsius to Kelvin is often necessary using the formula: \[ ext{K} = ext{°C} + 273.15\]For instance, \(331\,^{ ext{°C}}\) becomes \(604.15\, ext{K}\).

Converting units ensures that the Ideal Gas Law formula provides accurate results under uniform measurement conditions.

Pressure Calculations

Calculating pressure is an integral part of solving gas-related problems. Pressure, noted as \(P\), can be computed using the rearranged Ideal Gas Law. Consider the relationship:\[P = \frac{nRT}{V}\]Here's how this calculation unfolds:

• Identify Required Values: Start with known values of moles \(n\), volume \(V\), temperature \(T\), and select the correct value for the gas constant \(R\) based on your pressure units.
• Apply the Formula: Substitute the known values into the equation to calculate pressure. Using \(R = 62.364 \, ext{L*mmHg/mol*K}\) for mmHg, you can solve for \(P\) in conditions such as:\[P = \frac{(0.421 \, ext{mol})(62.364 \, ext{L*mmHg/mol*K})(223 \, ext{K})}{43.0 imes 10^{-3} \, ext{L}}\]This calculation gives the pressure as approximately \(681.40 \, ext{mmHg}\).

Recognizing when and how to use these calculations is key to predicting gas behavior in different scenarios.

Mole Calculations

The number of moles \(n\) of a gas is a measure of how many molecules are present and can be determined using the Ideal Gas Law. Here's the process:\[n = \frac{PV}{RT}\]Here's a step-by-step guide to calculating moles:

• Isolate Variables: Using known values for pressure \(P\), volume \(V\), and temperature \(T\), along with the gas constant \(R\), set up your equation.
• Substitute and Solve: Input the variables into the equation to solve for \(n\). For example:\[n = \frac{(7.74 \, \times 10^{3} \, ext{Pa})(12.2 \, \times 10^{-3} \, ext{L})}{(8.314 \, rac{ ext{J}}{ ext{mol} ext{*K}})(298 \, ext{K})}\]This calculation results in approximately \(0.0040 \, ext{mol}\).

Understanding mole calculations helps in determining the composition of gases and their reactivity, playing a fundamental role in chemical equations and reactions.