One Cubic Meter Of An Ideal Gas At 600 K And 1000 Malek KPa Expands To Five Times Its Initial Volume
Understanding the behavior of gases under different conditions is fundamental in thermodynamics and engineering applications. When an ideal gas undergoes expansion, its pressure, volume, and temperature change according to well-established laws. In this article, we delve into a specific scenario where a gas initially occupies a volume of one cubic meter at a temperature of 600 K and a pressure of 1000 Malek KPa, and then expands to five times its original volume. We will analyze the process using principles such as the ideal gas law, thermodynamic equations, and practical implications.
Initial Conditions of the Gas
Before exploring the expansion process, it’s essential to understand the initial state of the gas.
Initial Volume (V₁)
- V₁ = 1 m³
Initial Temperature (T₁)
- T₁ = 600 K
Initial Pressure (P₁)
- P₁ = 1000 Malek KPa
Understanding the Units: Malek KPa
- Malek KPa is a unit of pressure. For standard calculations, it’s often necessary to convert to Pascals (Pa):
Therefore,
P₁ = 1000 Malek KPa = 1,000,000,000 Pa (or 1 GPa)
Note: The high pressure indicates a dense or high-pressure environment, which is typical in specialized industrial processes.
Applying the Ideal Gas Law
The behavior of ideal gases is governed by the ideal gas law:
\[ PV = nRT \]
Where:
- P = pressure
- V = volume
- n = number of moles
- R = universal gas constant (~8.314 J/(mol·K))
- T = temperature in Kelvin
Since the process involves a single sample and the gas is ideal, we assume n remains constant during the expansion.
Calculating the Number of Moles (n) Initially
Rearranged as:
\[ n = \frac{PV}{RT} \]
Plugging in the initial conditions:
\[ n = \frac{(1 \times 10^9\, \text{Pa}) \times 1\, \text{m}^3}{8.314\, \text{J/(mol·K)} \times 600\, \text{K}} \]
\[ n \approx \frac{1 \times 10^9}{8.314 \times 600} \]
\[ n \approx \frac{1 \times 10^9}{4988.4} \]
\[ n \approx 200,448\, \text{mol} \]
This is the amount of gas initially present.
The Expansion to Five Times the Initial Volume
The problem states that the gas expands to five times its original volume.
Final Volume (V₂)
- V₂ = 5 × V₁ = 5 m³
Analyzing the Expansion Process
The key question is: what is the nature of the process? Is it isothermal, adiabatic, or polytropic? The answer determines how pressure and temperature change during the expansion.
Assumption 1: Isothermal Expansion
If the expansion occurs at constant temperature (T₂ = T₁), then:
\[ PV = \text{constant} \]
So,
\[ P2 = \frac{nRT}{V2} \]
Given initial state:
\[ P1V1 = P2V2 \]
\[ P2 = P1 \times \frac{V1}{V2} = 1 \times 10^9\, \text{Pa} \times \frac{1}{5} = 2 \times 10^8\, \text{Pa} \]
Thus, the pressure drops to 200 MPa during expansion.
Temperature remains at 600 K in this idealized case.
Assumption 2: Adiabatic Expansion
If the process is adiabatic (no heat exchange), then:
\[ PV^\gamma = \text{constant} \]
Where:
- γ = cp/cv, the heat capacity ratio. For an ideal gas, γ depends on the specific gas. Typical values:
- Diatomic gases (like nitrogen, oxygen): γ ≈ 1.4
- Monatomic gases (like noble gases): γ ≈ 1.67
Assuming a diatomic ideal gas:
\[ P2 = P1 \times \left(\frac{V1}{V2}\right)^\gamma \]
\[ P_2 = 1 \times 10^9\, \text{Pa} \times \left(\frac{1}{5}\right)^{1.4} \]
Calculating:
\[ \left(\frac{1}{5}\right)^{1.4} \approx e^{1.4 \times \ln(1/5)} \]
\[ \ln(1/5) \approx -1.6094 \]
\[ 1.4 \times -1.6094 \approx -2.253 \]
\[ e^{-2.253} \approx 0.105 \]
Therefore,
\[ P_2 \approx 1 \times 10^9\, \text{Pa} \times 0.105 \approx 1.05 \times 10^8\, \text{Pa} \]
The pressure decreases significantly but not as much as in the isothermal case.
The temperature after expansion (T₂) can be found using the adiabatic relation:
\[ T2 = T1 \times \left(\frac{V1}{V2}\right)^{\gamma - 1} \]
\[ T_2 = 600\, \text{K} \times \left(\frac{1}{5}\right)^{0.4} \]
Calculate:
\[ \left(\frac{1}{5}\right)^{0.4} \approx e^{0.4 \times -1.6094} \approx e^{-0.6438} \approx 0.525 \]
So,
\[ T_2 \approx 600 \times 0.525 \approx 315\, \text{K} \]
In adiabatic expansion, the temperature drops from 600 K to approximately 315 K.
Implications of the Expansion
The nature of the process significantly influences the thermodynamic properties during expansion.
Energy Considerations
- Work Done by the Gas (W):
\[ W = \int{V1}^{V_2} P\, dV \]
- In isothermal process:
\[ W = nRT \ln \frac{V2}{V1} \]
\[ W = 200,448 \times 8.314 \times 600 \times \ln 5 \]
\[ W \approx 200,448 \times 8.314 \times 600 \times 1.6094 \]
\[ W \approx 200,448 \times 8.314 \times 600 \times 1.6094 \]
Calculate step-by-step:
- \( 8.314 \times 600 \approx 4988.4 \)
- \( 200,448 \times 4988.4 \approx 1.000 \times 10^{9} \) (roughly equal to initial PV)
- Multiply by 1.6094:
\[ W \approx 1 \times 10^{9} \times 1.6094 \approx 1.609 \times 10^{9} \text{J} \]
- In adiabatic process, the work done is less because the temperature drops, and the process involves both work and heat transfer.
Final State Properties
- Final Volume: 5 m³
- Final Pressure: Varies depending on the process (approximately 200 MPa in isothermal, 105 MPa in adiabatic)
- Final Temperature: 600 K (isothermal) or ~315 K (adiabatic)
Practical Applications and Real-World Relevance
Understanding such expansion processes has numerous applications across industries.
Industrial Gas Storage and Transportation
- High-pressure gases are stored in cylinders or tanks.
- Expansion processes are critical in designing safety protocols and calculating energy exchanges.
Power Generation and Thermodynamic Cycles
- Expansion of gases is fundamental in turbines and engines.
- The efficiency of turbines depends on adiabatic expansion and temperature drops.
Refrigeration and Air Conditioning
- Expansion valves operate similarly to controlled gas expansion, cooling the system.
Conclusion
The expansion of an
