Question

The figure below displays a closed cycle for a gas (the figure is not drawn to scale). The change in the internal energy of the gas as it moves from a to c along the path abc is -200 J. As it moves from c to d, 180 J must be transferred to it as heat. An additional transfer of 80 J to it as heat is needed as it moves from d to a. How much work is done on the gas as it moves from c to d?

How much work is done on the gas as it moves from c to d?

Answer 1

Internal energy only depends on the state, so as we go through the cycle and come back to a, there shouldn't be any change. From d to a, V did not change. Therefore all the heat went into increasing the internal energy.

Therefore, the increase in internal energy between c and d must be 200 - 80 = 120J

Since total of 180J heat was transferred in c-d process, 180-120=60J is the work done.

Answer: 60J

Answer 2

Using the first law of thermo we canwrite:     ΔE - W = Q   for any process. Now consider the tablebelow, which I'll use to sort out the given info:

                  ΔE             W            Q

A to C       -200 ?               ?

C to D         ?               ?               180

D to A         ?               0 80

Notice that I used 0 for work from D to A. This is because theprocess is constant volume, vertical line on the diagram, and thework is zero for a constant volume process.

Now, we can figure out a few of these ? by realizing that inthe last line we have W and Q... so we can getΔE!!   It must be 80, since   ΔE = W+ Q in any line!

Also, the total change in energy, i.e. the total ΔE, fora cyclic process (i.e. you get back to where you started, in thiscase that is point A) must be zero. Notice now that yourvalues of ΔE are  

     -200 ? 80        if these total tozero, then the ? in the C to D process must be 120

Great! Now our table looks like this:

                 ΔE             W            Q

A to C       -200 ?               ?

C toD         120 ?            180

D toA         80               0               80

Hey... notice now we can determine the work done from C toD!   Since ΔE - W = Q for the process, wehave:        120 - W = 180 or the work fromC to D is -60 Joules