I'm confused about what the units of $C_v$ for an ideal monoatomic gas are if we derive its value from the formula : $$ C_v = \frac{f}{2}R$$ where $f$ represents the degree of freedom at room temperature.

(Wondering if they are $\frac{L.atm}{mol.K}$ or $\frac{J}{mol.K}$ if we use the above formula).

I encountered this in a problem which asked the change in enthalpy for a given process ($\triangle H$) under irreversible conditions and I used the formula $\triangle H = nC_v\triangle T$ but got the units wrong . Thank you in advance!

  • $\begingroup$ Weird units for R. Have you had a look at the Wikipedia entry: en.wikipedia.org/wiki/Gas_constant $\endgroup$ Sep 17 at 20:39
  • $\begingroup$ @Martin-マーチン do you mean that its units depend on what units I've taken for R? $\endgroup$
    – CapHim YT
    Sep 18 at 2:26
  • $\begingroup$ @Martin-マーチン I have seen this unit for R only here on CH SE, multiple times, derived from R = pV/(nT) $\endgroup$
    – Poutnik
    Sep 18 at 5:13
  • $\begingroup$ If a tree has the height 5 m and if another tree is three times taller, can it have height 15 cm ? $\endgroup$
    – Poutnik
    Sep 18 at 5:28
  • $\begingroup$ For an ideal gas, $\Delta H=nC_p\Delta T$. However, if the process is irreversible, such the initial equilibrium pressure, the final equilibrium pressure, and the external pressure throughout the process are all not equal, then $\Delta H$ is not equal to Q. $\endgroup$ Sep 18 at 11:20


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