Question

Neon is compressed from 100 kPa and 24°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.

Answer 1

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:  

$PV = RT$  

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:

$V_{1}= \frac {RT}{P_{1}}$      

$V_{1}= \frac {0.4119 \frac{kPa\cdot m^{3}}{kg\cdot K} \cdot (24 + 273 K)}{100 kPa}$

$V_{1}= 1.22 \frac{m^{3}}{kg}$

Similarly, the V₂ is:

$V_{2}= \frac {RT}{P_{2}}$  

$V_{2}= \frac {0.4119 \frac{kPa\cdot m^{3}}{kg\cdot K} \cdot (24 + 273 K)}{500 kPa}$

$V_{2}= 0.25 \frac{m^{3}}{kg}$

Now, the change in the specific volume because the compressor is:

$V_{2} - V_{1} = 0.25 - 1.22 \frac{m^{3}}{kg}$

$V_{2} - V_{1} = -0.97 \frac{m^{3}}{kg}$  

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:

$\Delta H = C_{p} \cdot \Delta T$    

$\Delta H = 1.0299 \frac{kJ}{kg \cdot K} \cdot 0$    

$\Delta H = 0 \frac{kJ}{kg}$

Have a nice day!