# What happens when throttling in an isothermal process

Consider an isothermal expansion of an ideal gas (always at ambient temperature $$T_0$$), from some initial pressure $$p_i$$ to the ambient pressure $$p_0$$, such that the maximum work available is:

$$W = mRT_0\ln(\frac{p_i}{p_0}) = p_iV_i\ln(\frac{p_i}{p_0})$$. (1)

If the pressure is throttled to some intermediate pressure, $$p_1, then presumably the expansion work can be be split into two phases, one with constant pressure $$p_1$$ and one with variable pressure, i.e. $$p. the maximum work available should then be the sum of these two phases, i.e.

$$p_1V_1\ln(\frac{p_1}{p_0})$$ (variable) + $$V_i(P_i-P_1)$$ (constant) (2)

For an ideal gas, the exergy is then just $$mRT\ln(p_i/p_0)$$, so without the throttling the exergy efficiency is 100%.

My question is, firstly, is this correct? Secondly, is the exergy destruction in the valve (1)-(2)? Thirdly, can you throttle (regulate pressure) without destroying a significant amount of exergy?

Furthermore, how do you draw ideal gas throttling on a PV diagram?

• The internal energy of ideal fluids (liquids or gases) depends only on temperature. Work is path dependent and may take a reversible or irreversible path. Exergy is best understood as the irreversible work along any path. Throttling is an irreversible process, while an isothermal REVERSIBLE process is just what it says it is. Any pV process for an ideal gas is always shown as the proper combination of isobars, isochors, isotherms, or adiabates for an ideal gas. – Jeffrey J Weimer Feb 2 '19 at 18:50