Hi, I asked this question before, but wanted to check my solution is correct:
Question: A cylinder fitted with a piston has a volume of 0.1 m3 and contains 0.5 kg of steam at 0.4 MPa. Heat is transferred to the steam until the temperature is 300 °C, as a result the volume increases to three times and the change in internal energy is 680.1 kJ. Determine the work and heat transfer for this process. Neglect the change in kinetic and potential energies.
Solution: I don't think we're interested in tables, what do you think?
Work = P.dV = P(V2 – V1)
Also, V2 = V1 x 3 - So V2 = 0.3m3
Work = 400 kPa x (0.3 – 0.1)
W = 80 kJ
dQ = dU + dW
dQ = 680.1 + 80
dQ = 760.1 kJ
Hi,
No, W = P.dV can not be used for steam. Steam is mixture of vapor and liquid phase.(pure substance). Above equations are valid for ideal gases only.
As, the steam is mixture of vapor and liquid. They may change their phases depending on the pressure and temperature. Liquid may get converted into vapor and vice-versa.
In case of ideal gases, it remains in the gaseous phase and does not change phase.(Unless it is compressed to large extent)