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.

Short answer

(a) STP stands for Standard Temperature and Pressure, which are 0°C (273.15 K) and 1 atm (101.325 kPa or 760 mmHg). (b) The molar volume of an ideal gas at STP is 22.41 L/mol. (c) The molar volume of an ideal gas at 25°C and 1 atm pressure is 24.47 L/mol.

Step-by-step solution

(a) Definition of STP

STP stands for Standard Temperature and Pressure. It is a reference condition used in chemistry, physics, and engineering to simplify calculations involving gases. The conditions represented by STP are: - Temperature: 0°C (273.15 K) - Pressure: 1 atm (101.325 kPa or 760 mmHg)

(b) Molar volume of an ideal gas at STP

The ideal gas law formula is given by: \(PV=nRT\), where P is pressure, V is volume, n is the amount of gas in moles, R is the ideal gas constant, and T is temperature. At STP, we want to find the molar volume (volume per mole) of an ideal gas. We will use Avogadro's law, which states that equal volumes of gases at the same temperature and pressure contain an equal number of particles (moles). One mole of an ideal gas at STP occupies a volume called the molar volume. For an ideal gas: Molar volume (V) = \( \frac{RT}{P} \) For STP conditions, temperature T = 273.15 K and pressure P = 1 atm. The ideal gas constant R = 0.0821 L atm / K mol Now, we can calculate the molar volume: Molar volume (V) = \( \frac{(0.0821 \, L \, atm / K \,mol) (273.15 \, K)}{1 \, atm} \) Molar volume (V) = 22.41 L/mol So, the molar volume of an ideal gas at STP is 22.41 L/mol.

(c) Molar volume of an ideal gas at 25°C and 1 atm pressure

Now, we need to calculate the molar volume of an ideal gas at room temperature, which is given as 25°C (or 298.15 K), and at 1 atm pressure. We will use the same formula as in part (b), but this time with the temperature T = 298.15 K and the same pressure P = 1 atm : Molar volume (V) = \( \frac{RT}{P} \) Molar volume (V) = \( \frac{(0.0821\, L\, atm / K\, mol)(298.15\, K)}{1\, atm} \) Molar volume (V) = 24.47 L/mol So, the molar volume of an ideal gas at 25°C (room temperature) and 1 atm pressure is 24.47 L/mol.

Key concepts

Ideal Gas Law

The ideal gas law is an essential equation in chemistry that helps us understand the behavior of gases. It is expressed as \(PV = nRT\), where:

• \(P\) stands for pressure measured in atmospheres (atm).
• \(V\) is the volume of the gas in liters (L).
• \(n\) represents the number of moles of gas.
• \(R\) is the ideal gas constant, which typically has a value of 0.0821 L atm / K mol.
• \(T\) is the temperature in Kelvin (K).

This equation helps illustrate the relationship between the properties of an ideal gas. If you know any three of the variables, you can calculate the fourth. By applying this formula, we can determine the molar volume of gases under different conditions including STP and room temperature.

STP Conditions

STP, or Standard Temperature and Pressure, is a baseline set of conditions for comparing gas reactions across various scientific fields. These standardized conditions are crucial for making calculations simpler and ensuring consistency in experiments.
At STP, the temperature is precisely 0°C (which also translates to 273.15 K), and the pressure is 1 atm (or equivalently 101.325 kPa).
Under these conditions, one mole of an ideal gas occupies a volume known as the molar volume, which we can calculate using the ideal gas law formula. The formula for molar volume at STP is:\[V = \frac{RT}{P}\]Substituting in the values specific to STP conditions, with \(R = 0.0821 \text{ L atm / K mol}\), \(T = 273.15 \text{ K}\), and \(P = 1 \text{ atm}\), results in a molar volume of 22.41 L/mol. This means every mole of an ideal gas at STP takes up 22.41 liters of space.

Room Temperature Calculations

Room temperature is commonly accepted as approximately 25°C, which equals 298.15 K. Chemistry problems often involve calculating gas properties at this temperature because it reflects many real-world laboratory and environmental conditions.

To calculate the molar volume of an ideal gas at room temperature (25°C) and 1 atm pressure, we apply the ideal gas law equation as follows:\[Molar \ Volume \, (V) = \frac{RT}{P} \]Here, \(T\) is updated to 298.15 K, while \(P\) and \(R\) remain 1 atm and 0.0821 L atm / K mol, respectively. Plugging these values into the formula gives:\[Molar \ Volume \, (V) = \frac{(0.0821 \times 298.15)}{1} \approx 24.47 \text{ L/mol}\]As a result, at room temperature and 1 atm pressure, one mole of an ideal gas occupies about 24.47 liters. This slight increase from the STP molar volume reflects the impact of higher temperatures on gas expansion.