# How do you determine the heat transfer from a P-V diagram?

I doubt this question has been addressed properly before, but if there are similar answers, do direct them to me.

I am currently studying the First Law of Thermodynamics, which includes the p-V diagram and of course, $\Delta U = Q_{to} + W_{on}$.

My question is: how do I determine the heat supplied to the system from a given p-V diagram? I know $W_{on} = -p(V_2-V_1)$, but is there a way to find $Q_{to}$ for a given path?

• I guess what you want is to integrate over the constant volume heat capacity $C_V(T)$. – TMOTTM Jul 10 '13 at 7:03
In the following, $R$ is the gas constant, $\gamma$ is the heat capacity ratio of the gas ($and$n$is the number of moles. • AB-isothermal.$\Delta W=Q$; area under the curve depicts the work done , ie. heat intake.$\Delta W=nRT\ln\dfrac{V_2}{V_1}$• BC-isobaric. Can't be calculated directly from curve. otherwise use,$Q=\Delta W + \Delta U$where$\Delta W=P\Delta V$and$\Delta U = n{c_v}\Delta T = {R \over {\gamma - 1}}n\Delta T$• CD-isochoric.$\Delta W=0Q=\Delta U = m{c_v}\Delta T = {R \over {\gamma - 1}}m\Delta T$. • DA-adiabatic . No heat exchange. • for any other general curve use$\Delta U +\Delta W = Q$anyhow. • How to know which part of the curve represents which kind of process? It is clear that BC is isobaric and CD is isochoric but how to know about the nature of the processes represented by the curves AB and DA? AB may be literally any polytropic process which has an inverse dependence of P on V and so, it is not necessary that it be isothermal. – user106570 Sep 17 '16 at 6:27 If you have a path on$p-V$diagram that is parametrized by some parameter$x$, so that$p=p(x),\,V=V(x)$, then: $$dU=\delta Q+p(x)dV(x)$$ here$Q$is the total heat received by the system (it is negative if system releases heat). I write$\delta Q$to indicate that$Q$is not a function of state, and$\delta Q$is not a full differential. Assume now that we are dealing with an ideal gas that has$f$degrees of freedom per particle ($f=3$for temperatures that are much lower than rotational and oscillational excitation energies). Then$U$is given by: $$U=\frac{f}{2}NkT=\frac{f}{2}pV$$ So, we have: $$\delta Q=\frac{f}{2}V(x)dp(x)+\frac{f-2}{2}p(x)dV(x)=\left(\frac{f}{2}V(x)p'(x)+\frac{f-2}{2}p(x)V'(x)\right)dx$$ Then the total received heat is $$Q_{to}=\int_Adx\left(\frac{f}{2}V(x)p'(x)+\frac{f-2}{2}p(x)V'(x)\right)$$ where$A$is the set of values of parameter$x$where the integrand is positive: $$A=\left\{x\left|\left(\frac{f}{2}V(x)p'(x)+\frac{f-2}{2}p(x)V'(x)\right)>0\right.\right\}$$ It is possible to determine$Q_{to}$for an arbitrary process on$p-V$diagram, specific cases are discussed by 007. One way is to find out the internal change energy of the system and infer the heat transfer to the system from that and the work done: $$\delta Q_\text{to}=dU-\delta W_\text{on}.$$ If you have a handle on the system's entropy, on the other hand, then you can use the Gibbs relation, $$\delta Q_\text{to}=TdS,$$to find the heat delivered. In general, though, any method is going to require more information about the system, because for the same process (and thus the same work performed) systems with different entropies$S=S(U,V)$(or equivalently with different energy functions,$U=U(S,V)$) will heat up by different amounts. If you have a path on$p-V$diagram that is$p=F(V)$, then using $$dU=\delta Q-pdV \implies \delta Q=dU+pdV$$ NOTE MINUS SIGN as$pdV$is work done BY the system.$Q$is the total heat received by the system (it is negative if system releases heat). Assume we are dealing with an ideal gas with$f$degrees of freedom per particle ($f=3$for monatomic gas). Then$U$is given by: $$U=\frac{f}{2}nRT=\frac{f}{2}pV$$ So, $$dU=\frac{f}{2}(pdV+Vdp)$$ $$\delta Q=\frac{f+2}{2}pdV+\frac{f}{2}Vdp$$ then using$p=F(V)$and$dp=F'(V)dV$$$\delta Q=(\frac{f+2}{2}F(V)+\frac{fV}{2}V F'(V))dV$$ Which you can just integrate along the path, care has to be taken though for the sign of$\delta Q$along the path - if you are looking for total heat in instead of net heat be careful if$\delta Q\$ becomes negative along the path as that will just give net heat...