Rank in order, from largest to smallest, the amount of work W51toW54done by the gas in each of the cycles shown in FIGURE Q21.2. Explain.

Short Answer

Expert verified

Rank in order isW3>W2=W1>W4

Step by step solution

01

Step :1 Introduction 

The work done in thermodynamics is represented by the area under the PV diagram. In order for work to be done in a closed system, there must be some volume displacement. In other words, an isochoric process (volume = constant) produces no work.

02

Step :2 Explanation

When a closed loop is established in a clockwise manner, the quantity of work performed by the system equals the area enclosed by the loop.

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Most popular questions from this chapter

A heat engine with 0.20molof a monatomic ideal gas initially fills a2000cm3cylinder at 600K. The gas goes through the following closed cycle:

Isothermal expansion to 4000cm3.

Isochoric cooling to 300K.

Isothermal compression to2000cm3 .

Isochoric heating to 600K.

How much work does this engine do per cycle and what is its thermal efficiency?

FIGURE CP21.70shows two insulated compartments separated by a thin wall. The left side contains 0.060molof helium at an initial temperature of 600Kand the right side contains 0.030molof helium at an initial temperature of 300K. The compartment on the right is attached to a vertical cylinder, above which the air pressure is 1.0atm. A 10-cm-diameter,2.0kg piston can slide without friction up and down the cylinder. Neither the cylinder diameter nor the volumes of the compartments are known.

a. What is the final temperature?

b. How much heat is transferred from the left side to the right side?

c. How high is the piston lifted due to this heat transfer?

d. What fraction of the heat is converted into work?

A heat engine using a diatomic ideal gas goes through the following closed cycle:

  • Isothermal compression until the volume is halved.
  • Isobaric expansion until the volume is restored to its initial value.

Isochoric cooling until the pressure is restored to its initial value. What are the thermal efficiencies of (a) this heat engine and

(b) a Carnot engine operating between the highest and lowest temperatures reached by this engine?

A heat engine running backward is called a refrigerator if its purpose is to extract heat from a cold reservoir. The same engine running backward is called a heat pump if its purpose is to exhaust warm air into the hot reservoir. Heat pumps are widely used for home heating. You can think of a heat pump as a refrigerator that is cooling the already cold outdoors and, with its exhaust heat QH, warming the indoors. Perhaps this seems a little silly, but consider the following. Electricity can be directly used to heat a home by passing an electric current through a heating coil. This is a direct, 100%conversion of work to heat. That is, 15kWof electric power (generated by doing work at the rate of 15kJ/sat the power plant) produces heat energy inside the home at a rate of 15kJ/s. Suppose that the neighbor’s home has a heat pump with a coefficient of performance of 5.0, a realistic value. Note that “what you get” with a heat pump is heat delivered, QH, so a heat pump’s coefficient of performance is defined asK=QH/Win.

a. How much electric power inkWdoes the heat pump use to deliver 15kJ/sof heat energy to the house?

b. An average price for electricity is about 40MJper dollar. A furnace or heat pump will run typically 250 hours per month during the winter. What does one month’s heating cost in the home with a 15kW electric heater and in the home of the neighbor who uses a heat pump?

In Problems 65through 68you are given the equation(s) used to solve a problem. For each of these, you are to

a. Write a realistic problem for which this is the correct equation(s).

b. Finish the solution of the problem.

4.0=QC/Win

QH=100J

See all solutions

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