Chapter 20: Problem 4
An ideal gas undergoes an isothermal expansion. What will happen to its entropy? a) It will increase. c) It's impossible to determine. b) It will decrease. d) It will remain unchanged.
Chapter 20: Problem 4
An ideal gas undergoes an isothermal expansion. What will happen to its entropy? a) It will increase. c) It's impossible to determine. b) It will decrease. d) It will remain unchanged.
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Get started for freeAn Otto engine has a maximum efficiency of \(20.0 \%\) find the compression ratio. Assume that the gas is diatomic.
One of your friends begins to talk about how unfortunate the Second Law of Thermodynamics is, how sad it is that entropy must always increase, leading to the irreversible degradation of useful energy into heat and the decay of all things. Is there any counterargument you could give that would suggest that the Second Law is in fact a blessing?
The entropy of a macroscopic state is given by \(S=k_{B} \ln w\) where \(k_{\mathrm{B}}\) is the Boltzmann constant and \(w\) is the number of possible microscopic states. Calculate the change in entropy when \(n\) moles of an ideal gas undergo free expansion to fill the entire volume of a box after a barrier between the two halves of the box is removed.
Suppose a Brayton engine (see Problem 20.26 ) is run as a refrigerator. In this case, the cycle begins at temperature \(T_{1}\), and the gas is isobarically expanded until it reaches temperature \(T_{4}\). Then the gas is adiabatically compressed, until its temperature is \(T_{3}\). It is then isobarically compressed, and the temperature changes to \(T_{2}\). Finally, it is adiabatically expanded until it returns to temperature \(T_{1}\) - a) Sketch this cycle on a \(p V\) -diagram. b) Show that the coefficient of performance of the engine is given by \(K=\left(T_{4}-T_{1}\right) /\left(T_{3}-T_{2}-T_{4}+T_{1}\right) .\)
The change in entropy of a system can be calculated because a) it depends only on the c) entropy always increases. initial and final states. d) none of the above. b) any process is reversible.
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