# Obtaining the ration of $Q:\Delta U:W$ for isobaric process

I know the First Law of Thermodynamics:

$$dU = dQ-dW$$

and I also know the following relations:

$$C_p-C_v= R$$ for an isobaric process $$dQ = nC_p\Delta T$$

Now my textbook states this relation for an isobaric process (without any derivation):

$$Q:\Delta U:W= nC_p\Delta T:nC_v\Delta T:nR\Delta T = C_p:C_v:R$$

How do I derive this ratio? I couldn't find it's derivation on the net or in another book.

In a reversible isobaric process, you are adding heat to the gas Q and it is expanding at constant pressure P to do reversible work on its surroundings. So: $$W=P\Delta V$$

$$\Delta U=Q-W=Q-P\Delta V$$But, from the ideal gas law, $PV=nRT$, so $$W=nR\Delta T\tag{1}$$ and $$\Delta U=Q-nR\Delta T$$

So, $$Q=\Delta U+nR\Delta T$$ But for an ideal gas: $$\Delta U=nC_v\Delta T\tag{2}$$ So, $$Q=nC_v\Delta T+nR\Delta T=nR(C_v+R)\Delta T=nC_p\Delta T\tag{3}$$ Therefore, from Eons. 1-3, we have: $$Q:\Delta U:W=nC_p\Delta T:nC_v\Delta T:nR\Delta T$$

I'm showing you the derivation of $\Delta U=nCv\Delta T$.

Consider $n$ moles of an ideal gas kept in a rigid container at initial pressure $P$,volume $V$ and temperature $T$. Now $\Delta Q$ heat is supplied to gas.Its temperature rises from $T$ to $T+ \Delta T$ whereas volume remains constant.

Work done by gas, $W=0$. From first law of thermodynamics,

$\Delta Q_v=\Delta U$

Now you can easily find the ratio.

• This is an isochoric process, not an isobaric process. Commented Aug 16, 2017 at 20:11
• Yes I have derived it keeping volume constant. Commented Aug 18, 2017 at 4:05
• At constant volume, the W is zero. Commented Aug 18, 2017 at 4:10