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Chapter 15: 16Q (page 412)

Question:A gas is allowed to expand (a) adiabatically and (b) isothermally. In each process, does the entropy increase, decrease, or stay the same? Explain.

Short Answer

Expert verified

(a) In adiabatic process, the entropy will stay the same and (b) in isothermal process, the entropy will increase.

Step by step solution

01

Understanding adiabatic and isothermal process 

In the adiabatic process, the rate of heat transfer is equivalent to zero. On the other hand, in isothermal process the system temperature is always constant.

02

Explanation of change in entropy in adiabatic process 

When a certain gas expands adiabatically the heat transfer is equal to zero that is \(Q = 0\).

The relation of change in entropy is given by,

\(\Delta S = \frac{Q}{T}\)

Here, \(\Delta S\) is the change in entropy and T is the temperature.

On plugging the values in the above relation.

\(\begin{array}{l}\Delta S = \frac{0}{T}\\\Delta S = 0\end{array}\)

Thus, there will be no change in entropy that is \(\Delta S = 0\).

03

Explanation of change in entropy in isothermal process 

When a gas expands isothermally, the variation in its internal energy becomes zero. Also, the gas has to perform work on its surrounding. Hence, due to work done, there will be heat flow into the gas. The entropy of the gas relies on the heat flow so that the entropy will increase in this case.

Thus, there will be increase in entropy.

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Most popular questions from this chapter

Question:(II) A heat engine exhausts its heat at 340掳C and has a Carnot efficiency of 36%. What exhaust temperature would enable it to achieve a Carnot efficiency of 42%?

Question: (II) (a) What is the coefficient of performance of an ideal heat pump that extracts heat from 6掳C air outside and deposits heat inside a house at 24掳C? (b) If this heat pump operates on 1200 W of electrical power, what is the maximum heat it can deliver into the house each hour? See Problem 35.

Question: (III) The PV diagram in Fig. 15鈥�23 shows two possible states of a system containing 1.75 moles of a monatomic ideal gas. \(\left( {{P_1} = {P_2} = {\bf{425}}\;{{\bf{N}} \mathord{\left/{\vphantom {{\bf{N}} {{{\bf{m}}^{\bf{2}}}}}} \right.} {{{\bf{m}}^{\bf{2}}}}},\;{V_1} = {\bf{2}}{\bf{.00}}\;{{\bf{m}}^{\bf{3}}},\;{V_2} = {\bf{8}}{\bf{.00}}\;{{\bf{m}}^{\bf{3}}}.} \right)\) (a) Draw the process which depicts an isobaric expansion from state 1 to state 2, and label this process A. (b) Find the work done by the gas and the change in internal energy of the gas in process A. (c) Draw the two-step process which depicts an isothermal expansion from state 1 to the volume \({V_2}\), followed by an isovolumetric increase in temperature to state 2, and label this process B. (d) Find the change in internal energy of the gas for the two-step process B.

(II) Suppose that you repeatedly shake six coins in your hand and drop them on the floor. Construct a table showing the number of microstates that correspond to each macrostate. What is the probability of obtaining

(a) three heads and three tails, and

(b) six heads?

On a very hot day, could you cool your kitchen by leaving the refrigerator door open?

(a) Yes, but it would be very expensive.

(b) Yes, but only if the humidity is below 50%.

(c) No, the refrigerator would exhaust the same amount of heat into the room as it takes out of the room.

(d) No, the heat exhausted by the refrigerator into the room is more than the heat the refrigerator takes out of the room.
See all solutions

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