Hey guys, I'm working on calculating the enthalpy, entropy and internal energy of substances. treating them as an ideal gas. I wonder if I could be pointed in the right direction with some calculations. Enthalpy change: h_{2}-h_{1}=∫c_{p} dT+ ∫[v-T(∂v/∂T)_{P}]dP At ideal gas the pressure integral goes to zero. So we have just the integral of C_{p} with respect to T. All correct so far? Its with the entropy and internal energy where I get confused. ds = (C_{p})dT - (∂v/∂T)_{P} dP is there a way to simplify this further at ideal gas to make it easily calculated ? I guess internal energy can be calculated from h using h = u + Pv. Thanks very much Look forward to any replies. 1. The problem statement, all variables and given/known data 2. Relevant equations 3. The attempt at a solution
I am not sure how you get this. If H = U + PV, then dH = dU + PdV + VdP and ΔH = ΔU + ∫PdV + ∫VdP = ΔQ + ∫VdP So, if pressure is constant, ∫VdP = 0, so ΔH = ΔQ = ∫C_{p}dT Not in general. This is true only if P is constant. Since U and PV are state functions, ΔU and Δ(PV) are independent of the path between two states so they are the same whether the path is reversible or irreversible. For a reversible path, ΔH = ΔQ + ∫VdP = ∫TdS + ∫VdP. Consequently, this must be true for all paths. So you can use: dH = TdS + VdP AM