Compute the specific heat capacity of ideal gas under constant $V$ and $p$ > Compute the specific heat capacities at constant volume and constant pressure for air at standard temperature and pressure, assuming it is diatomic ideal gas and a molecular mass of 28u. I have the equation for an ideal gas : $pV=Nk_BT$ and the caloric equation is : $E=\frac{5}{2}Nk_BT$ Now my definition for heat capacity is: $C=\frac{\delta Q}{dT}$. (Specific heat capacity will this divided by the mass). Now I get that $$ C=\frac{\delta Q}{dT}=\frac{dE+pdV}{dT}$$ and $$C_V=\left(\frac{\partial E}{\partial T} \right)_V \ \ \ \text{,} \ \ \ C_p=\left(\frac{\partial E}{\partial T} \right)_p+p\left(\frac{\partial V}{\partial T} \right)_p$$ I tried $C_V=\left(\frac{\partial E}{\partial T} \right)_V=\left(\frac{5}{2}Nk_B \right)_V$ but I cannot see how I can proceed with this. Nor do I think $C_p$ will be any easier to use

In physics it's $q=\Delta U+w$ and in chemistry it's $\Delta U=q+w$, don't know what it is in maths. but anyways I start with chemistry definition, since every such equation differs only by a sign.

At constant volume, $dV=0$, so $$\Delta W=-pdV=0$$
so $$\Delta Q=\Delta E=\left(\frac{5}{2}\right)Nk_B\Delta T$$
so $$C_v=\frac{\partial Q}{\partial T}=\left(\frac{5}{2}\right)Nk_B$$ At constant pressure, $$\Delta H=\Delta E+\Delta (pV)=\Delta Q+\Delta W+pdV=\Delta Q+(-pdV)+pdV=\Delta Q$$ so $$\Delta Q=\Delta H=\Delta E+\Delta(pV)=\left(\frac{5}{2}\right)Nk_B\Delta T+\underbrace{Nk_B\Delta T}_{\text{actually it's }\Delta(Nk_BT) }=\left(\frac{5}{2}+1\right)Nk_BT$$, so $$C_p=\frac{\partial Q}{\partial T}=\left(\frac{5}{2}+1\right)Nk_B$$