
For an isolated system, \[\Delta U\] = 0, what will be \[\Delta S\] ?
Answer
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Hint: There are a total of three types of systems, open, closed and isolated. In an open system, exchange of mass and energy both is allowed. In a closed system, only exchange of energy is allowed. In an isolated system, neither mass nor energy exchange is allowed, thus there will be no energy change in an isolated system.
Complete answer: The value of \[\Delta S\] in an isolated system will be positive.
The symbol \[\Delta U\] represents the change in internal energy of a system, and the symbol \[\Delta S\] represents the change in entropy of a system. Entropy represents the randomness of a system.
Now, here we have been given an isolated system, thus there is no exchange or transfer of mass and energy. Therefore, the change in internal energy ( \[\Delta U\] ) automatically becomes zero. As there is no change in internal energy of the system, the randomness of the molecules will increase. Thus, the entropy of an isolated system will increase, thus making the change in entropy ( \[\Delta S\] ) positive.
For example, in a thermos, there is no exchange of mass or energy which makes it an isolated system. Thus the change in entropy ( \[\Delta S\] ) inside the thermos will be positive and the change in internal energy ( \[\Delta U\] ) will be zero.
Let’s see another example, imagine there are two isolated systems connected with each other through a valve which is closed initially. There is air filled in both the containers, now as we open the valve, molecules will start moving with more randomness as there is more space to move without regularity. But, the internal energy of the whole system will not change and thus the change in internal energy ( \[\Delta U\] ) will be zero.
Note:
In open and closed systems, more regularity may be maintained which makes entropy go low, and thus makes the change in entropy negative. In an isolated system, regularity is not maintained as internal energy is zero, giving more space to the particles to move.
Complete answer: The value of \[\Delta S\] in an isolated system will be positive.
The symbol \[\Delta U\] represents the change in internal energy of a system, and the symbol \[\Delta S\] represents the change in entropy of a system. Entropy represents the randomness of a system.
Now, here we have been given an isolated system, thus there is no exchange or transfer of mass and energy. Therefore, the change in internal energy ( \[\Delta U\] ) automatically becomes zero. As there is no change in internal energy of the system, the randomness of the molecules will increase. Thus, the entropy of an isolated system will increase, thus making the change in entropy ( \[\Delta S\] ) positive.
For example, in a thermos, there is no exchange of mass or energy which makes it an isolated system. Thus the change in entropy ( \[\Delta S\] ) inside the thermos will be positive and the change in internal energy ( \[\Delta U\] ) will be zero.
Let’s see another example, imagine there are two isolated systems connected with each other through a valve which is closed initially. There is air filled in both the containers, now as we open the valve, molecules will start moving with more randomness as there is more space to move without regularity. But, the internal energy of the whole system will not change and thus the change in internal energy ( \[\Delta U\] ) will be zero.
Note:
In open and closed systems, more regularity may be maintained which makes entropy go low, and thus makes the change in entropy negative. In an isolated system, regularity is not maintained as internal energy is zero, giving more space to the particles to move.
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