# Problem involving P-V diagram

issacnewton
Hi

Here is a small problem. I already got the answer for part a. I have question about the part b.
For finding the heat energy input/output in each process from A to A, can't we use the
formula for the specific heat of the gas.

$$Q=m\, c \, (\Delta T)$$

But if we use that , I get $Q_{net} =0$ for the whole cycle.

That could not be correct, since I am getting net work done as

$$W_{net}=-4P_i V_i$$

Since in a thermodynamic cycle, we must have

$$\Delta E_{int} = 0$$

we deduce that

$$Q_{net}=-W_{net} = 4 P_i V_i$$

So what's wrong with my beginning approach to Qnet ?

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Staff Emeritus
Homework Helper
Hi

Here is a small problem. I already got the answer for part a. I have question about the part b.
For finding the heat energy input/output in each process from A to A, can't we use the
formula for the specific heat of the gas.

$$Q=m\, c \, (\Delta T)$$

But if we use that , I get $Q_{net} =0$ for the whole cycle.
This is a good question! Without thinking through the details, I suspect the answer lies in the fact that for some paths cv is relevant, and for others cp is used. So adding up c???ΔT for the four paths doesn't result in zero, even if adding up the ΔT's results in zero.

issacnewton
Hi Redbelly, yes , you are right. Gases would have different c values for constant volume and constant pressure.

issacnewton
Redbelly

I got it right now. Here are different heats I got.

$$Q_{AB}=mc_v (2T_A)$$

$$Q_{BC}=mc_p(6T_A)$$

$$Q_{CD}=mc_v(-6T_A)$$

$$Q_{DA}=mc_p(-2T_A)$$

so we got

$$Q_{net}=4mT_A(c_p-c_v)$$

But for the ideal gas, we have the relation between specific heats of the gas (with repect
to mass , not molar specific heats )

$$c_p-c_v = \frac{nR}{m}$$

also at point A, we have

$$P_i V_i = nRT_A$$

using these two relations we get

$$Q_{net}=4mT_A\left(\frac{nR}{m}\right)=4(nRT_A) = 4P_i V_i$$

$$Q_{net} = - W$$

which is what I got earlier. So it makes perfect sense now. Thanks for the hint