Question

Determine the temperature, in \(\mathrm{K}\), of \(5 \mathrm{~kg}\) of air at a pressure of \(0.3 \mathrm{MPa}\) and a volume of \(2.2 \mathrm{~m}^{3} .\) Verify that ideal gas behavior can be assumed for air under these conditions.

Short answer

The temperature is approximately 459.37 K. Ideal gas behavior can be assumed as Z is approximately 1.

Step-by-step solution

Identify the Ideal Gas Law

The Ideal Gas Law is given by \[ PV = nRT \], 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.

Convert Units if Necessary

Convert the given pressure into pascals (Pa):\[ 0.3 \,\mathrm{MPa} = 0.3 \times 10^6 \, \mathrm{Pa} = 300000 \, \mathrm{Pa} \].The volume is already given in cubic meters, so no conversion is needed for it.

Calculate the Number of Moles

Use the molar mass of air to find the number of moles. The molar mass of air is approximately 29 g/mol (0.029 kg/mol). The number of moles \(n\) is calculated by: \[ n = \frac{mass}{molar\, mass} = \frac{5 \, \mathrm{kg}}{0.029 \, \mathrm{kg/mol}} \approx 172.41 \, \mathrm{mol} \].

Solve for Temperature

Rearrange the Ideal Gas Law to solve for temperature \(T\): \[ T = \frac{PV}{nR} \], where \(R\) is the ideal gas constant \(8.314 \, \mathrm{J/(mol \, K)} \). Substituting the values: \[ T = \frac{(300000 \, \mathrm{Pa})(2.2 \, \mathrm{m^3})}{(172.41 \, \mathrm{mol})(8.314 \, \mathrm{J/(mol \, K)})} \approx 459.37 \, \mathrm{K} \].

Verify Ideal Gas Behavior

To verify whether air behaves as an ideal gas under these conditions, calculate the compressibility factor (Z). For ideal gases, Z approximately equals 1. Calculate \[ Z = \frac{PV}{nRT} \], and check if it's close to 1. Inserting the known values from above: \[ Z = \frac{(300000 \, \mathrm{Pa})(2.2 \, \mathrm{m^3})}{(172.41 \, \mathrm{mol})(8.314 \, \mathrm{J/(mol \, K)})(459.37 \, \mathrm{K})} \approx 1 \]. Therefore, the ideal gas behavior is assumed to be valid.

Key concepts

Pressure

Pressure is a measure of how much force is applied over a given area. It is an important concept in the Ideal Gas Law, where it is denoted by the symbol \(P\). Pressure is measured in units of pascals (Pa), which is equivalent to newtons per square meter (N/m²). In the exercise, the pressure is given as \(0.3 \, \text{MPa}\), which can be converted to pascals by multiplying by \(10^6\), yielding \(300000 \, \text{Pa}\). We need this value to apply the Ideal Gas Law accurately. High pressure means that gas particles are compressed into a smaller volume, increasing their interactions.

Volume

Volume is the amount of space that a substance occupies. In the context of the Ideal Gas Law, it is represented by the symbol \(V\) and measured in cubic meters (m³). For this exercise, the volume is given as \(2.2 \, \text{m}^3\). Volume and pressure are inversely related for a given amount of gas at a constant temperature; as one increases, the other decreases. When using the Ideal Gas Law, it’s vital to ensure that volume is in the correct units to avoid calculation errors.

Temperature

Temperature is a measure of the average kinetic energy of particles in a substance. When working with the Ideal Gas Law, temperature is denoted by \(T\) and must always be in Kelvin (K). Kelvin can be converted from Celsius by adding 273.15. In our problem, we calculated the temperature as approximately \(459.37 \, \text{K}\). Temperature is directly proportional to both pressure and volume when the number of moles is kept constant. Higher temperatures mean that gas particles move faster, resulting in greater pressure if the volume remains constant.

Molar Mass

Molar mass is the mass of one mole of a substance and is expressed in units of grams per mole (g/mol) or kilograms per mole (kg/mol). For air, the molar mass is approximately \(29 \, \text{g/mol}\) or \(0.029 \, \text{kg/mol}\). This value is crucial in calculating the number of moles, \(n\), by using the formula: \(n = \frac{mass}{\text{molar mass}}\). With a mass of \(5 \, \text{kg}\) of air, we calculated the number of moles as approximately \(172.41 \, \text{mol}\). Understanding molar mass helps in converting between the mass of a substance and the number of moles, making it easier to use in gas law equations.

Compressibility Factor

The compressibility factor, denoted by \(Z\), indicates how much a real gas deviates from ideal gas behavior. For ideal gases, Z is usually close to 1. To verify if air behaves like an ideal gas, we use the formula: \[ Z = \frac{PV}{nRT} \]. In our exercise, we calculated \(Z\) and found it to be approximately 1, meaning that air behaved almost like an ideal gas under the given conditions. Checking the compressibility factor helps in determining the accuracy and reliability of using the Ideal Gas Law for a given situation.