Homework Help: Thermodynamics problem really confusing

1. Oct 26, 2009

zarul

A irreversible heat engine operating between heat reservoirs at 600 K and 300 K has half the efficiency of a Carnot engine. Calculate the entropy change of the working fluid in the engine which is the system, the surroundings, and the universe per cycle of the engine if 1000 J are extracted from the hot reservoir per cycle.

I know the efficiency of the engine would be 1/2(1-300/600) which is equal to 1/4 but how do I relate this with the entropy change, I'm really confused. Please help out..
Thanks

2. Oct 26, 2009

Andrew Mason

First of all, since entropy is a state function (it does not depend on the path between two states) there can be no change in the entropy of the working fluid in a complete cycle since it returns to its original state.

Second, the surroundings are the reservoirs. This is also the rest of the universe as far as the engine is concerned.

The efficiency would indeed be .25 as you have found, which means:

$$\eta = W/Q_h = .25$$

Since 1000 J of heat is removed from the hot reservoir at 600 K then ____ J of heat is delivered to the cold reservoir at 300K. You can easily calculate the change of entropy using the definition:

$$dS = dQ/T$$

$$\Delta S$$ is the change in entropy of the hot reservoir (which is < 0) plus the change in entropy of the cold reservoir (which is > 0)

$$\Delta S = \int dQ_h/T_h + \int dQ_c/T_c$$

AM

3. Oct 26, 2009

zarul

Thank you Andrew for your quick reply..As for the entropy change for the whole cycle as zero, isn't it true only in the case of reversible cycle? Is it the same for IRREVERSIBLE cycle too. Once again thks for your input..

4. Oct 26, 2009

Andrew Mason

It does not matter. Entropy is a state function. (It can be thought of as the integral of dQ/T along the reversible path between two states). Since the working fluid is in exactly the same state at the end of the cycle as it began, there is no change in its entropy.

AM

5. Oct 26, 2009

zarul

I have one more problem if you don't mind.
Take the vapor pressures of pure benzene and pure toluene to be 103 and 32 mm Hg. resp. The pressure on a mixture of 1 mol benzene and 1 mol toluene is reduced until half of the mixture is vaporized. What is this pressure and what are the mole fractions of the benzene in the liquid and the vapor phases.

I am totally lost in this problem. I know we need to use Raoult's law and use the concept of Lever Law but I just can't connect the ideas to get the right answer.

Thanks

6. Oct 27, 2009

Andrew Mason

Zarul,

You will have to post this new problem in a separate thread. Welcome to PF, BTW.

AM