Is the formula for heat always true in Ideal gases?

Question

I was recently studying thermodynamics where I was stuck at a point which was not digestible to me.
In my notes there were two formulas written as $$ Q=nC_p\Delta{T}..(1) $$ $$ U=nC_v\Delta{T}..(2) $$ These are the equations for the respective quantities for Ideal gases.
Now my question is related to equation (1).I know that equation(2) is true in general and I have proved it through kinetic theory of gases, but My question is that ,does in general Equation (1) hold?
And if it does the from the first law of thermodynamics, we know that $$ dU=dQ-dW $$ By substitution we can say that $$ W=nR\Delta{T} $$ Which means that work done is independent of process it is taken through(Which must be wrong).I was having a doubt just because whenever my friend are solving questions they just apply this equation and still get the result(They don't care what is getting applied and just care about answer).Can someone please clarify my doubt.It will be really helpful.

Answer 1

I am of the type to always remind students to be careful, and not just use the terms "heat" and "work", because they are rather debated when you deal with more complicated topics later on.

Instead, your Equation (1) is better understood to be about enthalpy $$\mathrm dH=nC_p\,\mathrm dT\tag{1'}+\left[V-T\left(\frac{\partial V}{\partial T}\right)_p\right]\mathrm dp+\mu\,\mathrm dn$$ where you can see what you want when you pick a process whereby $\mathrm dp=0=\mathrm dn$

Instead of working with $$\mathrm dU=đQ-đW$$ It is less ambiguous to work with the well-defined stuff, the reversibly defined $$\mathrm dU=T\,\mathrm dS-p\,\mathrm dV+\mu\,\mathrm dn$$ and using $H=U+pV$ to get $$\mathrm dH=T\,\mathrm dS+V\mathrm dp+\mu\mathrm dn$$ and do the same manipulations to convert the $T\,\mathrm dS\to nC_x\,\mathrm dT$

Needless to say, all of these things do, in general, depend upon the specific path taken, and so you are correct, and your friends are very likely to face problems in the future if they do not pay attention to the details. However, because of how we have focused upon reversibility, quite a lot of these relations can be found just by changes in state functions, i.e. are path independent. That would partly go to explain why your friends could sloppily get the correct answer.