The Cancer Cure Posted December 11, 2009 Share Posted December 11, 2009 I'm sorry for this second question, but I am very lost in this chapter. I have no idea what to do on this problem. Please advise me. gas pressures N2 245 torr (1 L) Ne 745 torr (1 L) H2 489 torr (.5 L) Each of the bulbs contains a gas at the pressure shown. What is the pressure of the system when all the stopcocks are opened, assuming that the temperature remains constant? (We can neglect the volume of the capillary tubing connecting the bulbs.) Thanks guys. Link to comment Share on other sites More sharing options...
toastywombel Posted December 11, 2009 Share Posted December 11, 2009 Well when the stopcock is opened I would imagine that gas would escape and the pressure would be 0, assuming that the gas has somewhere to escape to and that its not sucked out. Then the pressure would be negative. Link to comment Share on other sites More sharing options...
hermanntrude Posted December 11, 2009 Share Posted December 11, 2009 yuck. I've seen questions like this before with two bulbs but not three. nasty question. however, what you need is a modified version of boyle's law (notice the numbers of moles and the temperature is constant. that should always scream "boyle" to you). basically the sum of the products of the pressures and volumes in the three separate flasks should equal the product of the pressure and volume of the single container after the stopcocks are opened: [math](P_{1}V_{1})+(P_{2}V_{2})+(P_{3}V_{3}) = P_{final}V_{final}[/math] where the subscipts 1,2 and 3 represent the values for the three flasks. Merged post follows: Consecutive posts mergedWell when the stopcock is opened I would imagine that gas would escape and the pressure would be 0, assuming that the gas has somewhere to escape to and that its not sucked out. Then the pressure would be negative. the stopcocks are between the flasks, i suspect. Also there's no such thing as a negative pressure. if the flasks were opened to the atmosphere, the final pressure would be atmospheric pressure. Link to comment Share on other sites More sharing options...
The Cancer Cure Posted December 11, 2009 Author Share Posted December 11, 2009 So it should be 245 + 745 + 244.5 = 1234.5 with three significant figures = 1240 torr? I tried 1240 torr (I'm using WebAssign by the way) and it gave me feedback that the answer was incorrect. What do you think is wrong? Link to comment Share on other sites More sharing options...
toastywombel Posted December 11, 2009 Share Posted December 11, 2009 yuck. I've seen questions like this before with two bulbs but not three. nasty question. however, what you need is a modified version of boyle's law (notice the numbers of moles and the temperature is constant. that should always scream "boyle" to you). basically the sum of the products of the pressures and volumes in the three separate flasks should equal the product of the pressure and volume of the single container after the stopcocks are opened: [math](P_{1}V_{1})+(P_{2}V_{2})+(P_{3}V_{3}) = P_{final}V_{final}[/math] where the subscipts 1,2 and 3 represent the values for the three flasks. Merged post follows: Consecutive posts merged the stopcocks are between the flasks, i suspect. Also there's no such thing as a negative pressure. if the flasks were opened to the atmosphere, the final pressure would be atmospheric pressure. Yeah I really was taking a random guess, I had to look up what a stopcock was haha. And thats why you will find me in the physics section most of the time ! I hope you guys get this problem rocked though! Good Luck! Link to comment Share on other sites More sharing options...
hermanntrude Posted December 11, 2009 Share Posted December 11, 2009 So it should be 245 + 745 + 244.5 = 1234.5 with three significant figures = 1240 torr? I tried 1240 torr (I'm using WebAssign by the way) and it gave me feedback that the answer was incorrect. What do you think is wrong? 1234.5 is Pfinal x Vfinal. Divide by the volume of the final system 1 Link to comment Share on other sites More sharing options...
The Cancer Cure Posted December 11, 2009 Author Share Posted December 11, 2009 Thank you so much! I'd keep clicking reputation but it won't let me. Link to comment Share on other sites More sharing options...
hermanntrude Posted December 11, 2009 Share Posted December 11, 2009 lol no problem! My students have their final exam tomorrow and I wish at least one of them cared half as much as you seem to. Link to comment Share on other sites More sharing options...
Phi for All Posted December 11, 2009 Share Posted December 11, 2009 I'd keep clicking reputation but it won't let me. We don't give hermanntrude enough recognition for the great work he does here, so I'll give him some more for you. Link to comment Share on other sites More sharing options...
hermanntrude Posted December 11, 2009 Share Posted December 11, 2009 We don't give hermanntrude enough recognition for the great work he does here, so I'll give him some more for you. thanks man! Link to comment Share on other sites More sharing options...
The Cancer Cure Posted December 11, 2009 Author Share Posted December 11, 2009 We don't give hermanntrude enough recognition for the great work he does here, so I'll give him some more for you. Thank you. :-p 1 Link to comment Share on other sites More sharing options...
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