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In engineering thermodynamics, gases are often encountered in various processes such as power generation, refrigeration, and propulsion. To analyze these processes effectively, it is essential to understand how gases behave under different conditions. The ideal gas model provides a simplified yet powerful way to describe gas behavior using mathematical equations. These equations relate pressure, volume, temperature, and the amount of gas, allowing engineers to predict and calculate gas properties in many practical situations.
Central to these calculations are the gas constants, which serve as fixed values that simplify the equations and make them universally applicable. Understanding these constants and the ideal gas equations is fundamental for solving problems in thermodynamics, especially in competitive exams where clarity and speed are crucial.
An ideal gas is a hypothetical gas that perfectly follows a set of assumptions, making its behavior predictable and mathematically simple. These assumptions are:
Because of these assumptions, the gas molecules do not interact except by bouncing off each other or the container walls, and the gas occupies the entire volume uniformly.
From experimental observations and kinetic theory, it was found that the pressure \(P\), volume \(V\), temperature \(T\), and amount of gas (in moles \(n\)) are related by the ideal gas equation of state:
Here, temperature must be measured on an absolute scale (Kelvin) because the gas behavior depends on the absolute thermal energy of molecules.
Why does this equation matter? It allows us to predict how a gas will respond if we change its volume, temperature, or pressure, which is essential in designing engines, compressors, and many other engineering systems.
The universal gas constant \(R\) is a fixed value that applies to all ideal gases and has the value:
\(R = 8.314 \, \text{J/mol·K}\)
This constant relates the energy per mole per degree Kelvin.
However, in many engineering problems, it is more convenient to work with the mass of the gas rather than the number of moles. For this, we use the specific gas constant \(R_s\), which is related to \(R\) by the molar mass \(M\) of the gas:
Since molar mass \(M\) varies for different gases, \(R_s\) is unique to each gas.
| Gas | Molar Mass \(M\) (kg/mol) | Specific Gas Constant \(R_s = \frac{R}{M}\) (J/kg·K) |
|---|---|---|
| Air (approx.) | 0.029 | 287 |
| Oxygen (O₂) | 0.032 | 259 |
| Nitrogen (N₂) | 0.028 | 296 |
Note: Air is a mixture of gases, but for many calculations, it is treated as a single gas with average molar mass.
Using the universal and specific gas constants, the ideal gas equation can be written in two common forms:
To convert between moles and mass, use the relation:
It is crucial to maintain consistent units throughout calculations:
Using consistent SI units ensures correct and reliable results.
Step 1: Identify known values:
Step 2: Use the ideal gas equation in mass form:
\[ P = \frac{m R_s T}{V} \]
Step 3: Substitute values:
\[ P = \frac{2 \times 287 \times 300}{0.5} = \frac{172200}{0.5} = 344400\, \text{Pa} \]
Step 4: Convert pressure to kilopascals (kPa):
\[ P = \frac{344400}{1000} = 344.4\, \text{kPa} \]
Answer: The pressure of the air is 344.4 kPa.
Step 1: Calculate the specific gas constant \(R_s\) for oxygen:
\[ R_s = \frac{R}{M} = \frac{8.314}{0.032} = 259.81\, \text{J/kg·K} \]
Step 2: Convert pressure to pascals:
\[ P = 200\, \text{kPa} = 200 \times 1000 = 200000\, \text{Pa} \]
Step 3: Use the ideal gas equation in mass form and solve for temperature \(T\):
\[ PV = m R_s T \implies T = \frac{PV}{m R_s} \]
Step 4: Substitute known values:
\[ T = \frac{200000 \times 1.5}{3 \times 259.81} = \frac{300000}{779.43} = 384.9\, \text{K} \]
Answer: The temperature of the oxygen gas is approximately 385 K.
Step 1: Use the relation:
\[ R_s = \frac{R}{M} \]
Step 2: Substitute values:
\[ R_s = \frac{8.314}{0.028} = 296.93\, \text{J/kg·K} \]
Answer: The specific gas constant for nitrogen is approximately 297 J/kg·K.
Step 1: Since the process is isothermal, temperature \(T\) remains constant.
Step 2: Use the ideal gas law for initial and final states:
\[ P_1 V_1 = P_2 V_2 \]
Step 3: Rearrange to find \(P_2\):
\[ P_2 = \frac{P_1 V_1}{V_2} \]
Step 4: Substitute known values:
\[ P_2 = \frac{100 \times 0.2}{0.1} = 200\, \text{kPa} \]
Answer: The final pressure after isothermal compression is 200 kPa.
Step 1: Convert volume to cubic meters:
\[ V = 500\, \text{liters} = 500 \times 10^{-3} = 0.5\, \text{m}^3 \]
Step 2: Convert temperature to kelvin:
\[ T = 27 + 273.15 = 300.15\, \text{K} \]
Step 3: Convert pressure to pascals:
\[ P = 2\, \text{atm} = 2 \times 101325 = 202650\, \text{Pa} \]
Step 4: Use ideal gas equation to verify or calculate missing variable (if mass or moles known). Here, the problem asks for pressure conversion, so final pressure is:
Answer: The pressure in pascals is 202650 Pa.
When to use: Whenever temperature is given in Celsius or other units.
When to use: When mass is given instead of number of moles.
When to use: To quickly calculate specific gas constants during exams.
When to use: Always, to avoid calculation errors.
When to use: When problem involves both moles and mass.
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