Problem Statement
Calculate the pressure exerted by 5 kg of nitrogen gas at a temperature of 10°C if the volume is 0.4 m\(^3\). Molecular weight of nitrogen is 28. Assume, ideal gas laws are applicable.
Given Data
- Mass of nitrogen, \(m = 5 \, \text{kg}\)
- Temperature, \(t = 10^\circ\text{C}\)
- Volume of nitrogen, \(V = 0.4 \, \text{m}^3\)
- Molecular weight of nitrogen, \(M_{mol} = 28 \, \text{kg/kmol}\)
Convert temperature to Kelvin: \(T = 10 + 273 = 283 \, \text{K}\)
Solution
1. Determine the Characteristic Gas Constant for Nitrogen (R)
The universal gas constant (\(R_u\)) is approximately \(8314 \, \text{Nm/(kmol} \cdot \text{K)}\). The characteristic gas constant (\(R\)) for nitrogen can be found by dividing the universal gas constant by the molecular weight of nitrogen:
2. Apply the Ideal Gas Law
The ideal gas law is given by the equation:
Where:
- \(P\) = Pressure (N/m\(^2\))
- \(V\) = Volume (m\(^3\))
- \(m\) = Mass (kg)
- \(R\) = Characteristic gas constant (Nm/(kg \(\cdot\) K))
- \(T\) = Absolute temperature (K)
We need to find \(P\):
$$ P = \frac{mRT}{V} $$ $$ P = \frac{5 \, \text{kg} \times 296.93 \, \text{Nm/(kg} \cdot \text{K)} \times 283 \, \text{K}}{0.4 \, \text{m}^3} $$ $$ P = \frac{419820.45}{0.4} \, \text{N/m}^2 $$ $$ P \approx 1049551.125 \, \text{N/m}^2 $$3. Convert Pressure to N/mm\(^2\)
Since \(1 \, \text{m}^2 = (1000 \, \text{mm})^2 = 10^6 \, \text{mm}^2\), then \(1 \, \text{N/m}^2 = 10^{-6} \, \text{N/mm}^2\).
The pressure exerted by the nitrogen gas is approximately \(1.05 \, \text{N/mm}^2\).
Explanation
1. Ideal Gas Law:
The problem utilizes the ideal gas law, \(PV = mRT\), which describes the behavior of ideal gases. This law is fundamental in thermodynamics and fluid mechanics for gases at relatively low pressures and high temperatures, where intermolecular forces and molecular volume are negligible.
2. Characteristic Gas Constant:
For a specific gas, the characteristic gas constant (\(R\)) is derived from the universal gas constant (\(R_u\)) and the gas's molecular weight. This constant is unique to each gas and allows the ideal gas law to be applied directly using the mass of the gas rather than the number of moles.
3. Unit Conversion:
Careful attention to units is essential. Temperatures must be in Kelvin for ideal gas law calculations. Pressure is initially calculated in N/m\(^2\) (Pascals) and then converted to N/mm\(^2\) for consistency with common engineering units.
Physical Meaning
1. Gas Pressure:
The calculated pressure represents the force exerted by the nitrogen gas per unit area on the walls of the cylinder. This pressure arises from the constant collisions of gas molecules with the container walls.
2. Relationship between Variables:
The ideal gas law demonstrates the direct relationship between pressure, mass, temperature, and the inverse relationship with volume. For a given mass and volume, increasing the temperature of the gas will increase its pressure, as the molecules move faster and collide more frequently and forcefully with the container walls.
3. Importance in Engineering:
Understanding gas pressure calculations is vital in many engineering fields, including the design of pressure vessels, pipelines, pneumatic systems, and internal combustion engines. It allows engineers to predict how gases will behave under different conditions and to ensure the safe and efficient operation of systems involving gases.
