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, with a standard temperature of \(0^{\circ}\mathrm{C}\) (273.15 K) and a standard pressure of 1 atm (101325 Pa). (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 room temperature (\(25^{\circ}\mathrm{C}\) or 298.15 K) and 1 atm pressure is approximately 24.5 L/mol.

Step-by-step solution

Part (a): Define STP

STP stands for Standard Temperature and Pressure. It is a reference value used in various fields of science for comparing measurements. The standard temperature is \(0^{\circ}\mathrm{C}\) (273.15 K), and the standard pressure is 1 atm (101325 Pa).

Part (b): Calculate the molar volume of an ideal gas at STP

To calculate the molar volume of an ideal gas, we use the Ideal Gas Law: \(PV=nRT\), where P is the pressure, V is the volume, n is the number of moles, R is the ideal gas constant and T is the temperature. Since we are interested in finding the molar volume, we can rewrite this as \(\frac{V}{n} = \frac{RT}{P}\). At STP, \(T=273.15\,\mathrm{K}\) and \(P=1\,\mathrm{atm}\). To calculate the molar volume, we take R to be in the units of L atm/mol K, where \(R=0.0821\,\mathrm{L\,atm/K\,mol}\). We then plug in the values and calculate molar volume: \[\frac{V}{n} = \frac{(0.0821\,\mathrm{L\,atm/K\,mol})(273.15\,\mathrm{K})}{1\,\mathrm{atm}} = 22.4\,\mathrm{L/mol}\] The molar volume of an ideal gas at STP is approximately 22.4 L/mol.

Part (c): Calculate the molar volume of an ideal gas at room temperature and 1 atm pressure

We are given that the temperature is \(25^{\circ}\mathrm{C}\) (298.15 K) and the pressure is 1 atm. We will again use the Ideal Gas Law to find the molar volume. The number of moles (n) remains as 1 mol, and the ideal gas constant (R) is the same as previously, \(R=0.0821\,\mathrm{L\,atm/K\,mol}\). We can now plug the values into the ideal gas equation: \[\frac{V}{n} = \frac{(0.0821\,\mathrm{L\,atm/K\,mol})(298.15\,\mathrm{K})}{1\,\mathrm{atm}} = 24.5\,\mathrm{L/mol}\] The molar volume of an ideal gas at room temperature (25°C) and 1 atm is approximately 24.5 L/mol.