Question 4.13: Problem: Oxygen contained in a cylinder is compressed adiaba...
Problem: Oxygen contained in a cylinder is compressed adiabatically by a piston from an initial state with V_1 = 0.2 m³ , P_1 = 200 kPa and T_1 = 25 ° C to a final temperature T_2 = 200 ° C . Find the work done during this process.
Find: Work W done during compression.
Known: Initial volume V_1 = 0.2 m³, initial pressure P_1 = 200 kPa, initial temperature T_1 = 25 °C, final temperature T_2 = 200 °C, adiabatic process so Q_{12 } = 0.
Diagram:
Assumptions: Oxygen is an ideal gas, no change in KE and PE of system, constant specific heat.
Governing Equations:
Ideal gas equation PV = mRT
First Law Q_{12 } +W_{12}=\Delta U_{12}
Properties: Gas constant of oxygen R = 0.2598 kJ / kgK (Appendix 1).
As a first approximation, from Appendix 1 for oxygen at 25 °C, c_v = 0.662 kJ / kgK. But we have more accurate information in Appendix 4, where c_v is given as a function of temperature.
The average temperature for the process is T_{avg} = (T_1 + T_2) / 2 = (25 °C + 200 °C) / 2 = 112.5 °C = 385.65 K. From Appendix 4, at 350 K, c_v = 0.668 kJ / kgK and at 400 K, c_v = 0.681 kJ / kgK.
We can do a linear interpolation between these values to get
\frac{c_{v,385.65 \ K}-0.668 \ kJ/kgK}{385.65 \ K-350 \ K} =\frac{0.681 \ kJ/kgK-0.668 \ kJ/kgK}{400 \ K-350 \ K}
Solving,
c_{v,385.65 \ K} = 0.677 \ 27 \ kJ/kgK.
We will use this number, but note that the difference from the value at 25 °C is only about 2%.
For most engineering applications such a small difference does not matter. You will need to exercise judgment as to whether it is necessary to interpolate values.
