# Is delta H = nC_p dT only valid for ideal gas for an isothermal process?

The related question is:

$$10\ \mathrm g$$ of argon gas is compressed isothermally and reversibly at a temperature of $$27\ \mathrm{^\circ C}$$ from $$10\ \mathrm l$$ to $$5\ \mathrm l$$. Calculate enthalpy change ($$\Delta H$$) for this process $$R=2.0\ \mathrm{cal\ K^{-1}\ mol^{-1}}$$.

I know that $$\mathrm dU=\left(\frac{\partial U}{\partial T}\right)_V\,\mathrm dT+\left(\frac{\partial U}{\partial V}\right)_T\,\mathrm dV$$

In case of ideal gas it will be equal to $$nC\,\mathrm dT$$ but as it is not given it is ideal gas. I have no idea on how to proceed.

• $dH=nC_pdT$ is valid only for an ideal gas. Feb 10, 2023 at 13:09
• Look up in a thermo book how to determine the effect of pressure on enthalpy for a real gas. Feb 10, 2023 at 13:11
• @ChetMiller I know that but how to solve it then. Feb 10, 2023 at 15:07
• What is the equation for dH in terms of dT and dP? Do you know that equation? Feb 10, 2023 at 18:32
• Well, you could get a fairly good first iteration by using ideal gas law, then plug into your known equations to refine. OR you could of course locate an expression for dH as @ChetMiller suggest. Feb 10, 2023 at 20:05

This might not be the expected textbook answer. For most practical purposes, however, you can treat the gas in this example as ideal gas. The example is about a monoatomic gas, well above its critical point temperature $$(27\ \mathrm{^\circ C}\gg-122.46\ \mathrm{^\circ C})$$, well below its critical point pressure $$(0.62\ \mathrm{bar}\ll48.63\ \mathrm{bar})$$, and well below its critical point density $$(1.0\ \mathrm{g\ l^{-1}}\ll535.6\ \mathrm{g\ l^{-1}})$$; therefore, ideal gas behaviour may be safely assumed.
In case the ideal gas approximation is not good enough, you would usually not try to find a better equation for a real gas in the real world; you would use tabulated numerical values instead. For state functions like internal energy $$U$$ and enthalpy $$H$$, this approach works so well because they depend only on the current equilibrium thermodynamic state of the system and not on the path that the system has taken to reach that state.
You need a set of quantities that describe the equilibrium states of the system. Usually temperature $$T$$, pressure $$p$$, and volume $$V$$ are used to describe a state point of a gas. However, you can as well use temperature $$T$$, mass $$m$$, and volume $$V$$ from the question. You can directly calculate the density $$\rho$$ from the given values and then use the values for temperature $$T$$ and density $$\rho$$ to find the corresponding tabulated values for the specific enthalpy $$h$$ of argon. For the values in the following table, I used NIST Reference Fluid Thermodynamic and Transport Properties Database (REFPROP) – NIST Standard Reference Database 23, Version 9 with an arbitrary number of five significant digits. On this occasion, I also looked up the values for pressure $$p$$ and specific internal energy $$u$$; however, those are not needed for this example. Using the values for specific enthalpy $$h$$ and the given value for mass $$m$$, the enthalpy $$H$$ of the sample can be directly calculated.
$$\begin{array}{lllll} \hline \text{Quantity} & \text{Quantity} & \text{Unit} & \text{Value} \\ \text{name} & \text{symbol} & & \text{State 1} & \text{State 2} \\ \hline \text{mass} & m & \mathrm g & 10.000 & 10.000 \\ \text{temperature} & T & \mathrm{^\circ C} & 27.000 & 27.000 \\ \text{volume} & V & \mathrm l & 10.000 & 5.0000 \\ \text{density} & \rho & \mathrm{g\ l^{-1}} & 1.0000 & 2.0000 \\ \hline \text{pressure} & p & \mathrm{bar} & 0.62448 & 1.2485 \\ \text{specific internal energy} & u & \mathrm{J\ g^{-1}} & 93.602 & 93.508 \\ \text{specific enthalpy} & h & \mathrm{J\ g^{-1}} & 156.05 & 155.93 \\ \hline \text{enthalpy} & H & \mathrm{J} & 1560.5 & 1559.3 \\ \hline \end{array}$$
As expected for a gas with approximately ideal behaviour, the resulting enthalpy change is very small: $$\Delta H=1559.3\ \mathrm J-1560.5\ \mathrm J=-1.2\ \mathrm J$$