-1
$\begingroup$

According to the first law of thermodynamics, $u=q+w$, where $u$ is changing in internal energy, $q$ is heat liberated and $w$ is the work done in the process.

Now at constant volume, $w=0$, hence $u=q$. Since $q$ is $n\cdot C_v\cdot T$, where $n$ is the amount of substance in mole, $C_v$ is the molar heat capacity at constant volume and $T$ is the temperature change. $u$ comes out to be $n\cdot C_v\cdot T$. However, this expression is valid for all other cases too whether volume is constant or not.

Same is the case with $H=n\cdot C_p\cdot T$. Why so?

$\endgroup$
1
$\begingroup$

The change in internal energy should be written as $\Delta U=nC_v\Delta T$, not $nC_vT$. This equation is valid for any temperature change (irrespective of whether the volume or pressure changes) only for an ideal gas. The equation for the change in enthalpy should be $$\Delta H=\Delta U+\Delta (PV)$$ For an ideal gas, this equation reduces to $$\Delta H=nC_v\Delta T+nR\Delta T=nC_p\Delta T$$ This equation is valid for any temperature change only for an ideal gas.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.