Two different gases occupy the two bulbs shown here. Consider the process that occurs when the stopcock is opened, assuming the gases behave ideally. (d) How does the process affect the entropy of the surroundings?

Video transcript

Hello Everyone in this video, we're dealing with four different statements talking about this depiction right here. These statements talk about the entropy of the surroundings. So in this diagram here we see that we have two different bowls of two different ideal gasses and if we go ahead and take out the stop cock, then these two gasses will go ahead and mix with each other. So the delta us or the entropy of the surroundings is equal to the negative delta H. Was of our entropy of the system over our temperature denoted as T. So we see that the entropy of the surroundings is associated with the help of our system given to us by this relationship right over here. So for ideal gasses there's going to be no attractive or repulsive inter molecular forces with that in mind. Then mixing the mixing of our ideal gasses, which is what happens if we go ahead and take out the stop cock here would mean that the delta H. Or the entropy of a system is equal to zero. There's no change of entropy and this is of course we're assuming that the heat exchange occurs only between the two bulbs. So then with that being said out of all the four choices of our statements talking about the mixing of our deal gasses right over here, is that the surrounding or the entropy of the surroundings is not affected by our process. So this right here is going to be my final answer for this problem