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To understand the specific volume, we focus on how much space a specific amount of natural gas occupies. The specific volume, denoted as \( v \), is essentially the volume occupied by one kilogram of a substance. It is the reciprocal of density and is given in units of \( \text{m}^3/\text{kg} \).

In the given exercise, the specific volume is found by solving the thermodynamic equation: \[ p = \frac{\big( 5.18 \times 10^{-3} \big)T}{v - 0.002668} - \frac{8.91 \times 10^{-3}}{v^2} onumber\]This equation relates pressure, specific volume, and temperature. To find \( v \), you substitute the known values of pressure (100 bar) and temperature (255 K), and solve the equation algebraically.

After finding the specific volume, you can compute the total volume of the container using the relation:\( V_{total} = v \times \text{mass} onumber \).This equation helps calculate the space required for 1000 kg of natural gas.

Key points to remember:

• Specific volume is an intrinsic property of a substance.
• It is crucial for determining the storage requirements.

Isotherms are curves that represent the state of natural gas at constant temperatures. In the context of the given exercise, we need to plot pressure versus specific volume at different constant temperatures (250 K, 500 K, and 1000 K).

Here's how you can plot these isotherms:

• Keep temperature, \( T \), constant for each plot.
• Vary the specific volume, \( v \), over a range of values.
• Use the thermodynamic equation:\[ p = \frac{\big( 5.18 \times 10^{-3} \big)T}{v - 0.002668} - \frac{8.91 \times 10^{-3}}{v^2} onumber\] to calculate the corresponding pressures, \( p \).

These plots help visualize how the pressure changes with specific volume at different temperatures.

Important notes about isotherms:

• Lower temperatures have steeper isotherms.
• Higher temperatures usually result in flatter curves.

Understanding isotherms is crucial for designing and operating systems involving natural gas storage.

Thermodynamic equations are mathematical models that describe the state of a system. They relate thermodynamic properties such as pressure, temperature, and specific volume.

In the exercise provided, the equation\[ p = \frac{\big( 5.18 \times 10^{-3} \big)T}{v - 0.002668} - \frac{8.91 \times 10^{-3}}{v^2} onumber \]is used to describe the relationship between pressure, specific volume, and temperature for natural gas.

Key elements of this equation:

• The term \( \big( 5.18 \times 10^{-3} \big)T / (v - 0.002668) \) represents the portion of pressure due to temperature.
• The term \( \frac{8.91 \times 10^{-3}}{v^2} \) accounts for intermolecular forces within the gas.

Understanding thermodynamic equations helps you predict how natural gas behaves under different conditions. These equations are essential tools for engineers and scientists working with gases, ensuring that they can control and optimize processes safely and efficiently.

Remember, solving these equations typically involves algebraic manipulation and sometimes numerical methods due to their complexity.