Understanding P-V Diagrams: Work Done and Heat Input in a Closed Loop Process

[quietrain]

in a close loop P-V diagram,

dQ = dE + PdV right?

so if i want to get the net work down by the gas , i just need to find the Work done for each process right?

but my dE is always 0 since cyclic process,

so for constant volume,

WD = ∫ PdV = 0

for constant temperature,

WD = ∫ PdV = nrT∫ VdV = nrT ln V

for constant pressure,

WD = ∫ PdV = P(V₂-V₁)

so since dE is 0, heat input = work done by the gas right? since from First law,

heat in = change in internal energy + work done by the gas

so in calculating the work done by the gas, i am calculating the heat input right?

thanks!

 

[m.e.t.a.]

Your reasoning looks correct to me: for a cyclic process whereby a fixed quantity of ideal gas is returned to its initial (P, V) coordinates, the net change in internal energy is zero. $\Delta U = Q - W = 0$, so $Q = W$.

 

[Andrew Mason]

so since dE is 0, heat input = work done by the gas right? since from First law,

heat in = change in internal energy + work done by the gas

What do you mean by heat in? Qh or Qh-Qc?

$\Delta Q = W$. The net heat flow into the system = work done by the system. The net heat flow into the system is the heat flow from the hot reservoir (Qh) minus the heat flow to the cold reservoir (Qc).

So:

$$W = \Delta Q = Q_h-Q_c$$

AM

 

[quietrain]

ah i see thanks!