# Work done by a heat engine given thermodynamic cycle

A heat engine uses an ideal gas as its working substance, and is described by the following thermodynamic cycle; Isobaric expansion at $$4 \text{ atm}$$ from a volume of $$4~\text L$$ to $$8~\text L$$. Isochoric pressure reduction to $$1\text{ atm}$$. Isobaric compression to half the volume. Isochoric pressure increase to $$4\text{ atm}$$. How much work does the heat machine produce per cycle?

I'm solving some previous exams in an introductory physics class. One of the chapters deals with simple thermodynamics. I'm struggling to really connect the theory when it comes to thermodynamic processes to actual calculations. The solution to the problem is simply $$W = 3 \cdot 4 = 12 \text{ atm}\cdot \text L = 1.2 \text{ kJ}$$.

I don't really understand anything about the thought process in how to calculate this though. I would very much appreciate an explanation as to how they arrive at these numbers?

More specifically; what is the work done and what happens in the 4 different stages, here?
1) "Isobaric expansion at $$4 \text{ atm}$$ from a volume of $$4~\text L$$ to $$8~\text L$$"
2) "Isochoric pressure reduction to $$1 \text{ atm}$$"
3) "Isobaric compression to half the volume"
4) "Isochoric pressure increase to $$4 \text{ atm}$$"

• What do you understand by the terms "isobaric" and "isochoric" ? – user207455 Apr 26 '19 at 10:32

HINT: This problem is just using the formula $$W=P\Delta V$$.

For an isochoric process, $$V$$ is constant, so $$W=0$$.

For an isobaric process, $$P$$ is constant, not $$V$$. So, $$W \not =0$$.

Net work done is $$W_{net}=W_1 +W_2 ...$$, where $$1,2 ...$$ are the steps in the process - in your case, 4 steps to complete a single cycle.

Pressure volume work is defined as

$$W=\int PdV$$

Isobaric means constant pressure so P come out of the integral and you have

$$W=P\Delta V$$

Note that expansion work is positive and compression is negative. Add the two isobaric work processes algebraically.

Isochoric means constant volume so the work done on those two processes is zero.

Hope this helps