(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.

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)

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.

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.