Why Does Using Specific Heat in a P-V Cycle Yield Zero Net Heat Transfer?

[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 Q_net ?

 

[Redbelly98]

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 c_v is relevant, and for others c_p 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