Adiabatic Compression Work Analysis for a Monoatomic Ideal Gas

This problem involves determining the work of compression for a monoatomic ideal gas undergoing an adiabatic process. Understanding the properties of ideal gases and the principles of adiabatic processes is crucial for solving such thermodynamic problems. We are given the initial and final pressures, initial temperature, specific heat ratio, and molecular weight of the gas.

Gas Properties and Initial Conditions

Let's list the given parameters for the monoatomic ideal gas:

• Initial Pressure (\(P_1\)): 0.1 MPa = 100 kPa
• Final Pressure (\(P_2\)): 0.2 MPa = 200 kPa
• Initial Temperature (\(T_1\)): 300 K
• Specific Heat Ratio (\(\gamma\)): 1.67 (for a monoatomic ideal gas, \(\gamma = 5/3 \approx 1.667\))
• Molecular Weight (M): 40 kg/kmol
• Universal Gas Constant (\(R_u\)): 8.314 KJ/kmol.K (assuming the common unit for universal gas constant as KJ/kmol.K, which is equivalent to kJ/(mol·K) when divided by 1000)
• Given relation: \((2)^{(\gamma-1)/\gamma}\) = 1.32

Specific Gas Constant Calculation

First, we need to calculate the specific gas constant (\(R\)) for the given monoatomic ideal gas. The specific gas constant is obtained by dividing the universal gas constant (\(R_u\)) by the molecular weight (M) of the gas.

The formula for the specific gas constant is:

\(R = \frac{R_u}{M}\)

Substituting the given values:

\(R = \frac{8.314 \text{ KJ/kmol.K}}{40 \text{ kg/kmol}}\)

\(R = 0.20785 \text{ KJ/kg.K}\)

Final Temperature Determination in Adiabatic Compression

For an adiabatic process, the relationship between temperature and pressure is given by:

\(\frac{T_2}{T_1} = \left(\frac{P_2}{P_1}\right)^{(\gamma-1)/\gamma}\)

We are given \(T_1 = 300 \text{ K}\), \(P_1 = 0.1 \text{ MPa}\), and \(P_2 = 0.2 \text{ MPa}\). We also have the specific heat ratio \(\gamma = 1.67\).

Substitute the pressure values into the equation:

\(\frac{T_2}{300 \text{ K}} = \left(\frac{0.2 \text{ MPa}}{0.1 \text{ MPa}}\right)^{(1.67-1)/1.67}\)

\(\frac{T_2}{300 \text{ K}} = (2)^{0.67/1.67}\)

The problem conveniently provides the value for \((2)^{(\gamma-1)/\gamma}\) as 1.32. This simplifies the calculation:

\(\frac{T_2}{300 \text{ K}} = 1.32\)

Now, calculate the final temperature \(T_2\):

\(T_2 = 300 \text{ K} \times 1.32\)

\(T_2 = 396 \text{ K}\)

Work of Compression Calculation

For an adiabatic process, the work done on the gas (work of compression) per unit mass can be calculated using the formula:

\(W_{\text{compression}} = \frac{R(T_2 - T_1)}{\gamma - 1}\)

Here, \(R\) is the specific gas constant, \(T_1\) is the initial temperature, \(T_2\) is the final temperature, and \(\gamma\) is the specific heat ratio.

Substitute the calculated and given values into the formula:

\(W_{\text{compression}} = \frac{0.20785 \text{ KJ/kg.K} \times (396 \text{ K} - 300 \text{ K})}{1.67 - 1}\)

\(W_{\text{compression}} = \frac{0.20785 \text{ KJ/kg.K} \times 96 \text{ K}}{0.67}\)

\(W_{\text{compression}} = \frac{19.9536 \text{ KJ/kg}}{0.67}\)

\(W_{\text{compression}} \approx 29.78149 \text{ KJ/kg}\)

Rounding this value, the work of compression is approximately 29.7 KJ/kg.

Summary of Results

The key steps and results are summarized in the table below:

The calculated work of compression is approximately 29.7 KJ/kg, which aligns with one of the provided options.