If we assume the ideal gas law $E=\frac{f}{2}pV$, and then differentiate get $$dE=\frac{f}{2}dpV+\frac{f}{2}pdV$$ do $\frac{f}{2}dpV$ stands for $dQ$, and $\frac{f}{2}pdV$ for $dW$ in first law of thermodynamics?
I just started thermodynamics is my understanding correct?
It’s not quite correct. I’ll note internal energy $U$ rather than $E$. It is true for any process: $$ dU=\delta W +\delta Q=-pdV+TdS $$ It is only for reversible processes that you can identify the terms namely: $$ \delta W= -pdV \\ \delta Q=TdS $$ Note the sign and overall factor that differs from your expression of work. In general, when considering internal energy, the natural variables to work with are volume and entropy i.e. $V,S$.
Your reasoning would work if you had rather said for a reversible process: $$ \delta W=-pdV \\ \delta Q= \frac{f+2}{2}pdV+\frac{f}{2}Vdp $$ The second equation is actually consistent with the formula of entropy of an ideal gas and the formula for heat.
Hope this helps.
