(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 \(/ \mathrm{mol}-\mathrm{K}\).

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.

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: \(V = 22.4 \mathrm{L/mol}\)

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: \(V = 24.5 \mathrm{L/mol}\)

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: \(R' = 0.0831 \mathrm{L \cdot bar / mol \cdot K}\)