A closed, nonconducting, horizontal cylinder is fitted with a nonconducting, frictionless, floating piston that divides the cylinder into Sections A and B. The two sections contain equal masses of air, initially at the same conditions, = 300 K and = 1(atm). An electrical heating element in Section A is activated, and the air temperatures slowly increase: in Section A because of heat transfer, and in Section A because of heat transfer, and

src="https://sciemce.com/media/4/7bc03e91afc40c5.png" class="w-image" /> in Section A because of heat transfer, and be the number of moles of air in Section A. For the process as described, evaluate one of the following sets of quantities:



What will be an ideal response?

(a) and if P(final) = 1.25(atm)

(b) and P (final), if = 425 K

(c) and P (final), if = 325 K

(d) and P (final), if

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We want to develop relationships between p, and . Then, given any one of the quantities, we can find the other three. One relationship is that the total volume remains constant:

We can rewrite this in terms of the number of moles in each chamber (), the pressure (which is the same in both chambers) and the temperature in each chamber:



Another relationship arises from the fact that the compression in chamber B is adiabatic, so



The third relationship derives from the energy balances on the two closed systems. If we let be the work done by the gas in chamber A on the gas in chamber B, then



and

Writing the internal energy changes in terms of the heat capacity and temperature changes:



or



Now, we can apply these three relationships to solve the problems posed in parts (a) through (d). You only had to do one part of your choice.

a. If we know the final pressure, then we can first use:



from which = 319.75 K

Next, we can use



to get



from which = 430.3 K

Finally, then,

b. This is a bit trickier. One way to approach it is to substitute into to get

We can then guess values of until we find one that gives the specified value of = 425/300 = 1.417. If

we want to iterate on the equation, we might write it as:



From part (a), we know that will be close to 1.25. So, we might iterate on the above starting from a guess of = 1.25.

This quickly converges to = 1.24, or p = 1.24 *1 atm = 1.24 atm.

Now, we can use this directly to get = 319.0 K.

Finally, the two temperatures are used to get

c. This one is a bit more straightforward using our equations. First, we can use to get = 1.3233 atm.

Then, we can use

to get



from which = 469 K.

Finally, the two temperatures are used to get

d. This time, we have to start with from which Substituting this into gives , or just = 744.3/300 = 2.481, from which p = 1.2405 atm. happened to conveniently cancel out.

Then we can use

to get

Finally, = 744.3 – = 425.2 K.

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