course PHY 122 For the assignment 13 experiment I don't have any of the connectors or the terminators that are required for the experiment. Could I come by Monday and pick some up?Thanks £œ{›©€ÀÌý„¹Ï¹£Ñ¢Û©`Éò¯üq¹®ÆÜ®É
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14:28:11 query problem 15 introductory problem sets temperature and volume information find final temperature. When temperature and volume remain constant what ratio remains constant?
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RESPONSE --> nR/P will also remain constant
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14:28:27 ** PV = n R T so n R / P = T / V; since T and V remain constant T / V and therefore n R / P remain constant; since R is constant it follows that n / P remains constant. **
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RESPONSE --> ok
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14:30:38 why, in terms of the ideal gas law, is T / V constant when only temperature and volume change?
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RESPONSE --> Because they are the only two variable on that side of the equation. When it is rearranged to V/T=nR/P
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14:33:07 ** STUDENT ANSWER AND INSTRUCTOR RESPONSE: They are inverselt proportional. They must change together to maintain that proportion. INSTRUCTOR RESPONSE: You haven't justified your answer in terms of the ideal gas law: PV = n R T so V / T = n R / P. If only T and V change, n and P don't change so n R / P is constant. Therefore V / T is constant, and so therefore is T / V. You need to be able to justify any of the less general relationships (Boyle's Law, Charles' Law, etc.) in terms of the general gas law. **
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RESPONSE --> OK, I see what you wher asking now. I think I was on the right track with your answer.
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14:58:32 prin and gen problem 14.3: 2500 Cal per day is how many Joules? At a dime per kilowatt hour, how much would this cost?
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RESPONSE --> The 2500 calories are dietary calories. and they are 1000 times a normal calorie.or a( kcal) 1kcal = 4.186*10^3J, so 2500kcal =10465000J = 10.5*10^6J =2.91kw/hr @ a rate of $ .10 per kw/hr It would cost about $.29 a day. or 29cents.
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15:03:00 One Cal (with capital C) is about 4200 Joules, so 2500 Cal is about 4200 * 2500 Joules = 13,000,000 Joules. A watt is a Joule per second, a kilowatt is 1000 Joules / second and a kiliowatt-hour is 1000 Joules / second * 3600 seconds = 3,600,000 Joules. 13,000,000 Joules / (3,600,000 Joules / kwh) = 4 kwh, rounded to the nearest whole kwh. This is about 40 cents worth of electricity. It's worth noting that you use 85% of this energy just keeping yourself warm, so the total amount of physical work you can produce in a day is worth less than a dime.
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RESPONSE --> "