Questions: Problem 04.053 - Change of internal energy for neon Neon is compressed from 100 kPa and 16.00°C to 500 kPa in an isothermal compressor. Determine the change in the specific volume and specific enthalpy of neon caused by this compression. The gas constant of neon is R=0.4119 kJ / kg · K, and the constant-pressure specific heat of neon is 1.0299 kJ / kg · K. (Include a minus sign if necessary.) The change in the specific volume is m^3 / kg. (Round the final answer to four decimal places.) The change in the specific enthalpy is kJ / kg.

Solution Steps

The problem states that neon is compressed isothermally from an initial pressure \( P_1 = 100 \, \text{kPa} \) and temperature \( T = 16.00^\circ \text{C} \) to a final pressure \( P_2 = 500 \, \text{kPa} \). First, convert the temperature to Kelvin:

\[ T = 16.00 + 273.15 = 289.15 \, \text{K} \]

Using the ideal gas law, the specific volume \( v_1 \) at the initial state can be calculated as:

\[ v_1 = \frac{R \cdot T}{P_1} \]

Substitute the given values:

\[ v_1 = \frac{0.4119 \, \text{kJ/kg} \cdot \text{K} \times 289.15 \, \text{K}}{100 \, \text{kPa}} = \frac{0.4119 \times 289.15}{100} \, \text{m}^3/\text{kg} \]

\[ v_1 = 1.1911 \, \text{m}^3/\text{kg} \]

Since the process is isothermal, the specific volume \( v_2 \) at the final state can be calculated using the relation:

\[ v_2 = \frac{R \cdot T}{P_2} \]

Substitute the given values:

\[ v_2 = \frac{0.4119 \, \text{kJ/kg} \cdot \text{K} \times 289.15 \, \text{K}}{500 \, \text{kPa}} = \frac{0.4119 \times 289.15}{500} \, \text{m}^3/\text{kg} \]

\[ v_2 = 0.2382 \, \text{m}^3/\text{kg} \]

The change in specific volume \( \Delta v \) is:

\[ \Delta v = v_2 - v_1 = 0.2382 - 1.1911 = -0.9529 \, \text{m}^3/\text{kg} \]

For an isothermal process, the change in specific enthalpy \( \Delta h \) for an ideal gas is zero because enthalpy is a function of temperature only, and the temperature remains constant.

\[ \Delta h = 0 \, \text{kJ/kg} \]

Final Answer