Answer:
ΔV = -0.97 m³/ kg
ΔH = 0 kJ/ kg
Explanation:
To determine the change in the specific volume we need to use the ideal gas law:
[tex] PV = RT [/tex]
where P: pressure of the gas V: volume of the gas, R: ideal gas constant= 0.4119 kJ/kg.K = 0.4119 kPa.m³/kg.K and T: temperature of the gas.
The V₁, at a compressed pressure is:
[tex] V_{1}= \frac {RT}{P_{1}} [/tex]
[tex] V_{1}= \frac {0.4119 \frac{kPa\cdot m^{3}}{kg\cdot K} \cdot (24 + 273 K)}{100 kPa} [/tex]
[tex] V_{1}= 1.22 \frac{m^{3}}{kg} [/tex]
Similarly, the V₂ is:
[tex] V_{2}= \frac {RT}{P_{2}} [/tex]
[tex] V_{2}= \frac {0.4119 \frac{kPa\cdot m^{3}}{kg\cdot K} \cdot (24 + 273 K)}{500 kPa} [/tex]
[tex] V_{2}= 0.25 \frac{m^{3}}{kg} [/tex]
Now, the change in the specific volume because the compressor is:
[tex] V_{2} - V_{1} = 0.25 - 1.22 \frac{m^{3}}{kg}[/tex]
[tex] V_{2} - V_{1} = -0.97 \frac{m^{3}}{kg}[/tex]
Finally, to calculate the change in the specific enthalpy, we need to remember that neon is an ideal gas and that is an isothermal process:
[tex] \Delta H = C_{p} \cdot \Delta T [/tex]
[tex] \Delta H = 1.0299 \frac{kJ}{kg \cdot K} \cdot 0 [/tex]
[tex] \Delta H = 0 \frac{kJ}{kg} [/tex]
Have a nice day!
