-2
$\begingroup$

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!

Improve this question
$\endgroup$
12
  • $\begingroup$ Weird units for R. Have you had a look at the Wikipedia entry: en.wikipedia.org/wiki/Gas_constant $\endgroup$
    – Martin - マーチン ♦
    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. $\endgroup$
    – Chet Miller
    Sep 18 at 11:20

0

Reset to default

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.