Calculate the density of air when the absolute pressure and the temperature are respectively \(140 \mathrm{kPa}\) and \(50{ }^{\circ} \mathrm{C}\) and \(\mathrm{R}=\) \(287 \mathrm{~J} \cdot \mathrm{kg}^{-1} \cdot \mathrm{K}^{-1}\)

The ideal gas law is a fundamental equation in chemistry and physics. It relates the pressure, volume, and temperature of a gas to the number of particles in it. The formula is:

\[ PV = nRT \]
Here:

\t
• P is the pressure of the gas
\t
• V is the volume
\t
• n is the number of moles
\t
• R is the gas constant
\t
• T is the absolute temperature

In the context of calculating the density of air, this equation helps us relate the pressure and temperature of the air to its density. By rearranging the equation to solve for density (\( \rho \) ), we get:

\[ \rho = \frac{P}{RT} \]

This shows that the density is directly proportional to pressure (\( P \) ) and inversely proportional to both the gas constant (\( R \) ) and temperature (\( T \) ). Understanding this relationship is key to solving problems involving the density of gases.

Converting temperature is a crucial step when using the ideal gas law. The formula requires the temperature to be in Kelvin (K), but temperatures are often given in Celsius (°C). To convert from Celsius to Kelvin, use the formula:

\[ T(K) = T(°C) + 273.15 \]
For example, if the temperature is 50°C:
\[ T(K) = 50 + 273.15 = 323.15 \text{ K} \]
This conversion ensures you're using the correct units for temperature in your calculations. It’s important to note that Kelvin starts at absolute zero, which is -273.15°C. This conversion will always add 273.15 to the Celsius temperature. Remembering this addition is key for accurately solving problems involving temperatures.

The density of air (\( \rho \) ) is a measure of how much mass is contained in a given volume. For gases, density can be calculated using the rearranged ideal gas law formula:

\[ \rho = \frac{P}{RT} \]
Where:

\t
• P is the absolute pressure
\t
• R is the gas constant
\t
• T is the absolute temperature in Kelvin

For instance, given the values:

\t
• P = 140 kPa = 140,000 Pa
\t
• R = 287 J/kg·K
\t
• T = 323.15 K

Inserting these values into the density formula, you perform the calculations to find:
\[ \rho = \frac{140,000}{287 \times 323.15} \ \rho ≈ 1.50 \text{ kg/m}^3 \]

This shows that at a pressure of 140 kPa and a temperature of 50°C, the density of air is approximately 1.50 kg/m³.

The relationship between pressure and temperature is central to understanding how gases behave. According to the ideal gas law, when the volume is constant, the pressure of a gas is directly proportional to its temperature. This can be expressed as:

\[ \frac{P1}{T1} = \frac{P2}{T2} \]
If the volume and the number of particles are constant, an increase in temperature will result in an increase in pressure, and vice versa. This is known as Gay-Lussac's law.

For example, if you increase the temperature of an air-filled balloon without changing its volume, the pressure inside the balloon will increase. When dealing with gases in the atmosphere, such as calculating air density, taking this relationship into account is vital. Understanding how pressure and temperature change together helps in accurately applying the ideal gas law.