Problem 29 - Chemistry The Central Science

Question

(a) What conditions are represented by the abbreviation STP? (b) What is the molar volume of an ideal gas at STP? (c) Room temperature is often assumed to be \(25^{\circ} \mathrm{C}\) . Calculate the molar volume of an ideal gas at \(25^{\circ} \mathrm{C}\) and 1 atm pressure. (d) If you measure pressure in bars instead of atmospheres, calculate the corresponding value of \(R\) in L-bar/mol-K.

Step-by-Step Solution

Answer

(a) STP stands for Standard Temperature and Pressure, which are \(0^{\circ}\mathrm{C} = 273.15\mathrm{K}\) and \(1 \mathrm{atm} = 101.325\mathrm{kPa}\). (b) The molar volume of an ideal gas at STP is approximately 22.4 L/mol. (c) The molar volume of an ideal gas at \(25^{\circ}\mathrm{C}\) and 1 atm pressure is approximately 24.5 L/mol. (d) The value of the gas constant R in L-bar/mol-K units is approximately 0.08314 L·bar/mol·K.

Step 1: (a) Conditions represented by STP

STP stands for Standard Temperature and Pressure. It represents a specific set of conditions, which are: - Temperature: \(0^{\circ}\mathrm{C} = 273.15\mathrm{K}\), and - Pressure: \(1 \mathrm{atm} = 101.325\mathrm{kPa}\).

Step 2: (b) Molar volume of an ideal gas at STP

To calculate the molar volume of an ideal gas at STP, we can use the Ideal Gas Law equation: \[PV = nRT\] where P is the pressure, V is the volume, n is the number of moles, R is the gas constant, and T is the temperature. We know that at STP, the pressure \(P = 1 \mathrm{atm}\), temperature \(T = 273.15\mathrm{K}\), and we will consider 1 mole of the gas, which means \(n = 1\). The gas constant \(R = 0.0821 \frac{\mathrm{L}\cdot\mathrm{atm}}{\mathrm{mol}\cdot\mathrm{K}}\). Now we can rearrange the Ideal Gas Law equation to solve for volume V: \[V = \frac{nRT}{P}\] Plugging in the known values: \[V = \frac{(1\,\mathrm{mol}) \times (0.0821 \frac{\mathrm{L}\cdot\mathrm{atm}}{\mathrm{mol}\cdot\mathrm{K}}) \times (273.15\,\mathrm{K})}{1\,\mathrm{atm}}\] The answer is: \[V \approx 22.4\,\mathrm{L/mol}\] So, the molar volume of an ideal gas at STP is approximately 22.4 L/mol.

Step 3: (c) Molar volume at 25°C and 1 atm pressure

To find the molar volume of an ideal gas at \(25^{\circ}\mathrm{C}\) and 1 atm pressure, we will follow the same steps as in part (b) but use the new temperature value. First, convert the given temperature to Kelvin: \(T = 25^{\circ}\mathrm{C} + 273.15 = 298.15\,\mathrm{K}\) Now, we can use the Ideal Gas Law equation with the new temperature value: \[V = \frac{nRT}{P}\] Plugging in the known values: \[V = \frac{(1\,\mathrm{mol}) \times (0.0821 \frac{\mathrm{L}\cdot\mathrm{atm}}{\mathrm{mol}\cdot\mathrm{K}}) \times (298.15\,\mathrm{K})}{1\,\mathrm{atm}}\] The answer is: \[V \approx 24.5\,\mathrm{L/mol}\] So, the molar volume of an ideal gas at \(25^{\circ}\mathrm{C}\) and 1 atm pressure is approximately 24.5 L/mol.

Step 4: (d) Value of R in L-bar/mol-K

In order to find the value of R in L-bar/mol-K units, we can use the following conversion factor: \(1\,\mathrm{atm} = 1.01325\,\mathrm{bar}\) We know that the gas constant, \(R = 0.0821 \frac{\mathrm{L}\cdot\mathrm{atm}}{\mathrm{mol}\cdot\mathrm{K}}\). Now, we can convert R to the desired units by multiplying the given value of R by the conversion factor: \[R_{\mathrm{L\cdot bar/mol\cdot K}} = R_{\mathrm{L\cdot atm/mol\cdot K}} \times \frac{1\,\mathrm{bar}}{1.01325\,\mathrm{atm}}\] \[R_{\mathrm{L\cdot bar/mol\cdot K}} = (0.0821 \frac{\mathrm{L}\cdot\mathrm{atm}}{\mathrm{mol}\cdot\mathrm{K}}) \times \frac{1\,\mathrm{bar}}{1.01325\,\mathrm{atm}}\] The answer is: \[R_{\mathrm{L\cdot bar/mol\cdot K}} \approx 0.08314\,\frac{\mathrm{L\cdot bar}}{\mathrm{mol\cdot K}}\] So, the value of the gas constant R in L-bar/mol-K units is approximately 0.08314 L·bar/mol·K.

Key Concepts

Standard Temperature and Pressure (STP)

Understanding Standard Temperature and Pressure (STP) is foundational in studying the behavior of gases. At STP, an ideal gas has specific properties that make calculations more straightforward. The terms Standard Temperature and Pressure refer to the condition of a gas at a temperature of 0 degrees Celsius (273.15 Kelvin) and a pressure of 1 atmosphere (101.325 kPa). These conditions are commonly used because they represent standard laboratory conditions, making it easier to compare experimental results. Molar volume is a critical concept associated with STP, which is the volume occupied by one mole of a gas. At STP, an ideal gas occupies 22.4 liters per mole, a figure derived from the Ideal Gas Law and empirical data.

For students grappling with these concepts, visualize STP as a benchmark or reference point for gas properties. This simplifies complex calculations and offers a standard against which changes in temperature or pressure can be measured and understood. When you work with gases, always consider whether the conditions are at STP to make accurate predictions and calculations.

Ideal Gas Law

The Ideal Gas Law is a fundamental equation that serves as a cornerstone for understanding the behavior of gases. It's formulated as PV = nRT, where P is the pressure, V is the volume, n is the number of moles, R is the gas constant, and T is the temperature in Kelvin. This equation aptly describes how pressure, volume, and temperature are interrelated for an ideal gas—a theoretical gas that perfectly follows the gas laws.

For students, mastering the Ideal Gas Law involves not just memorizing the equation, but truly comprehending how each variable impacts the others. For instance, if you were to increase the temperature (T) while keeping the number of moles (n) and pressure (P) constant, the volume (V) would also increase. Recognizing these relationships helps explain and predict gas behavior under various conditions. Real gases adhere closely to the Ideal Gas Law under many conditions, although they may deviate at high pressures or low temperatures where the assumptions of the ideal gas model break down.

Gas Constant (R) Value Conversion

In the realm of chemistry, the Gas Constant, often denoted as R, is a pivotal figure in the Ideal Gas Law. It links the physical properties of pressure, volume, temperature, and the amount of gas into a single equation. However, R can be expressed in several units, and converting between these units is crucial for solving problems accurately.

The gas constant's value varies based on the pressure and volume units used, such as liters and atmospheres, or bars and cubic meters. In the case of changing from atmospheres to bars, you would use the conversion factor of 1 atm equaling 1.01325 bars. Performing these conversions correctly ensures that your calculations remain consistent and dimensionally accurate, as mismatched units can lead to errors in the results. Whenever you're presented with R in unfamiliar units, always look for the necessary conversion factors to maintain the proper units throughout your calculations. Remember, paying attention to units can make or break a chemical calculation and understanding how to convert the gas constant properly allows for flexible application of the Ideal Gas Law across various scientific scenarios.