# On isentropic compression of ideal gases, work and internal energy

I'm trying to understand something about isentropic compression of Hydrogen. If you are assuming constant specific heats, the work done during an isentropic compression is

$$W = \frac{k}{k-1}\cdot RT_0\cdot\left(\left(p_\mathrm f/p_\mathrm i\right)^{0.285} - 1\right).$$

Let's start with hydrogen at $$p = 1\ \mathrm{bar}$$ at $$25\ \mathrm{^\circ C}$$ and go to $$p = 880\ \mathrm{bar}$$. If you plug in these values, and use $$k = 1.41$$, you come up with $$W \approx 26\,000\ \mathrm{kJ/kg}$$.

Now my question is, if this is an isentropic process (i.e. it's adiabatic) surely the work done is also directly changing internal energy? If I use the expression for temperature and pressure relations in an isentropic change, then you can find the temperature at the second stage to be $$T_{2} = T_{1}* 880^{0.285}$$ and from there, the temperature at the second stage to be $$= 2\,057\ \mathrm K$$.

Working from an ideal gas table for hydrogen, the change in internal energy is a bit different from the work expression evaluated; the change in internal energy found from the tables based on the change in temperatures is ~ 36000 kJ/kmol which does not equal this when converted to kJ/kg.

I just want to confirm that this is a valid way of arriving at the same answer – finding the isentropic temperature and then determining the change in internal energy, instead of plugging in the expression for the work.

...as an addendum, when I look at the enthalpy differences between these two pressures, it's possible that this does the trick because the enthalpy difference is 54,651 kJ/kmol (~26,000 kJ/kg for H2; there are some triangulation steps and approximations I'm leaving out because they really cannot be that significant) but I'm still curious to see if this is in general a correct approach for an exam.

• At 880 bar, I would not rely on ideal gas relationships. – Pieter Oct 14 at 10:57