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 \(101.3 \mathrm{kPa}\) pressure. \((\mathbf{d})\) If you measure pressure in bars instead of atmospheres, calculate the corresponding value of \(R\) in \(\mathrm{L}\) -bar \(/ \mathrm{mol}-\mathrm{K}\).

Short answer

(a) STP stands for Standard Temperature and Pressure, which represents a temperature of \(273.15 K\) and a pressure of \(1\) atm or \(101.3\) kPa. (b) The molar volume of an ideal gas at STP is \(22.4 \mathrm{L/mol}\). (c) The molar volume of an ideal gas at \(25^{\circ} \mathrm{C}\) and \(101.3\) kPa pressure is \(24.5 \mathrm{L/mol}\). (d) The value of R when measuring pressure in bars is \(0.0831 \mathrm{L \cdot bar / mol \cdot K}\).

Step-by-step solution

(a) Determine the conditions represented by STP

STP stands for Standard Temperature and Pressure. Under STP conditions, the temperature (T) is \(273.15 K\) and the pressure (P) is \(1\) atm or \(101.3\) kPa.

(b) Find the molar volume of an ideal gas at STP

We will use the ideal gas law, which is given by \(PV = nRT\), where P is pressure, V is volume, n is the number of moles, R is the universal gas constant, and T is the temperature. At STP, we have: - T = \(273.15 K\) - P = \(1\) atm = \(101.3\) kPa - For one mole of an ideal gas, n = \(1\) mol The value of R is \(0.0821 \mathrm{L} \cdot \mathrm{atm} / \mathrm{mol} \cdot K\) and we will find the volume V of the gas. So, the equation becomes: \(PV = RT\) Solving for V: \(V = \frac{RT}{P}\) Using the STP values: \(V = \frac{(0.0821 \mathrm{L} \cdot \mathrm{atm} / \mathrm{mol} \cdot K)(273.15 K)}{1 \mathrm{atm}}\) After the calculation, we get the molar volume of an ideal gas at STP as: \(V = 22.4 \mathrm{L/mol}\)

(c) Calculate the molar volume of an ideal gas at 25°C and 101.3 kPa pressure

For this question, make sure to convert the temperature to Kelvin (K) by adding \(273.15\): \(25^{\circ} \mathrm{C} + 273.15 = 298.15 K\) Now we have: - T = \(298.15 K\) - P = \(101.3\) kPa To find the volume V, use the ideal gas law again: \(PV = nRT\) Since we are asked for molar volume, we use n = \(1\) mol. When pressure is given in kPa, we need to use a different value for the universal gas constant: \(R = 8.314 \mathrm{L} \cdot \mathrm{kPa} / \mathrm{mol} \cdot K\) Using the given temperature and pressure, the equation becomes: \(V = \frac{(8.314 \mathrm{L} \cdot \mathrm{kPa} / \mathrm{mol} \cdot K)(298.15 K)}{101.3 \mathrm{kPa}}\) After the calculation, we get the molar volume of an ideal gas at \(25^{\circ} \mathrm{C}\) and \(101.3\) kPa pressure as: \(V = 24.5 \mathrm{L/mol}\)

(d) Calculate the value of R in L-bar/mol-K if the pressure is measured in bars

We know that \(1 \mathrm{bar} = 100 \mathrm{kPa}\), so we need to convert the units of the gas constant R accordingly. Currently, we are given the value of R for the units kPa as \(R = 8.314 \mathrm{L} \cdot \mathrm{kPa} / \mathrm{mol} \cdot K\) To convert to L-bar/mol-K, we use the conversion factor: \(R' = \frac{8.314 \mathrm{L} \cdot \mathrm{kPa}}{1 \mathrm{mol} \cdot K} \times \frac{1 \mathrm{bar}}{100 \mathrm{kPa}}\) After the calculation, we get the value of R for pressure measured in bars as: \(R' = 0.0831 \mathrm{L \cdot bar / mol \cdot K}\)

Key concepts

Standard Temperature and Pressure (STP)

When discussing gases in chemistry, we often refer to STP, which stands for Standard Temperature and Pressure. These conditions are commonly used as a reference to simplify calculations and express gas volumes.
At STP, the temperature is defined as exactly 273.15 Kelvin, equivalent to 0 degrees Celsius. This is the freezing point of water. The pressure at STP is standardized at 1 atmosphere, which is also equal to 101.3 kilopascals.

These precise conditions allow scientists to compare the behavior of gases under controlled circumstances. It's important to memorize these values, as they are frequently used in chemistry problems involving gases.

Molar Volume

Molar volume is a key concept in understanding how gases occupy space. It refers to the volume that one mole of any gas will occupy under specified conditions of temperature and pressure. For an ideal gas at STP, the molar volume is calculated to be 22.4 liters per mole.
This value is derived from the ideal gas law, represented as \(PV = nRT\), where:

• \(P\) stands for pressure.
• \(V\) is the volume.
• \(n\) denotes the number of moles.
• \(R\) is the universal gas constant.
• \(T\) symbolizes temperature, in Kelvin.

At STP, using the universal gas constant of 0.0821 L atm/mol K, we find that for one mole of gas, the volume \( V \) becomes 22.4 liters. This constant is useful when converting between the number of moles of a gas and the volume it occupies under standard conditions.

Universal Gas Constant

The universal gas constant, denoted as \(R\), is integral to the ideal gas law formula, \(PV = nRT\). It serves as a bridging constant that relates the conditions of gases to a standard context.
Depending on the units used for pressure and volume, \(R\) may have different numerical values.
Here are some commonly used values:

• When pressure is in atmospheres and volume in liters, \(R = 0.0821\) L atm/mol K.
• When pressure is in kilopascals and volume in liters, \(R = 8.314\) L kPa/mol K.
• If you measure pressure in bars, \(R\) takes the value of 0.0831 L bar/mol K.

These different values ensure we can apply the ideal gas law using whatever pressure units we are working with.
Understanding the correct value of \(R\) for your units is crucial for accurately applying the ideal gas equation.