Problem involving P-V diagram

• issacnewton
In summary, Redbelly says that in order to find the heat energy input/output in each process from A to A, one can use the formula for the specific heat of the gas, but this won't work if the gas has different c values for constant volume and constant pressure. Qnet will be negative for the whole cycle, since the work done (Wnet) will be negative for every process.

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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IssacNewton said:
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.

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

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