hdhdhd49jdhd
03.09.2019 •
Chemistry
(b) use the first law of thermodynamic to calculate au for the following situations: (i) a coiled spring unwinds producing 153 j of work and losing 37 j as heat to friction (1 mark) (ii) an insulated bucket of water is stirred by a paddle, which provides 289 j of work. (1 mark) (iii) adding 1 kj of heat to a sample of gas in an isovolumetric process. (1 mark) (iv) the isothermal compression of a gas from 26 l to 5 l. (1 mark) (v) the isobaric expansion of a gas from 5 l to 18 l at a constant pressure of 950 kpa caused by the addition of 15.6 kj of heat.
Solved
Show answers
More tips
- S Society and Politics Как правильно поддерживать температуру в квартире для здоровья...
- L Leisure and Entertainment How to Make a Crab Trap in Just a Few Minutes...
- H Health and Medicine How Much Does Abortion Cost? Expert Answers and Insights...
- S Sport How to Build Arm Muscles? Effective Exercises and Tips...
- H Health and Medicine When can it be said that a person has a normal pulse?...
- A Art and Culture When Will Eurovision 2011 Take Place?...
- S Style and Beauty How to Choose the Perfect Hair Straightener?...
- F Family and Home Why Having Pets at Home is Good for Your Health...
- H Health and Medicine How to perform artificial respiration?...
- H Health and Medicine 10 Tips for Avoiding Vitamin Deficiency...
Answers on questions: Chemistry
- B Biology A scientist studied human skin cells as they produced two genetically identical daughter cells from a single parent cells the scientist mapped the steps shown...
- B Business Windsor company purchased equipment for $206,400 on october 1, 2020. it is estimated that the equipment will have a useful life of 8 years and a salvage value of...
- P Physics Apanpipe is made of five pipes. the longest pipe is 25 centimeters long and the shortest is 5 centimeters long. which of these pipe produces the highest-frequency...
Ответ:
(i) ΔU = 116 J
(ii) ΔU = 289 J
(iii) ΔU = 1 KJ
(iv) ΔU = 0 J
(v) ΔU = 3.25 KJ
Explanation:
first law:
ΔU = Q + W(i) W = 153 J; Q = - 37 J ( Q ( - ), losing friction )
⇒ ΔU = 153 - 37 = 116 J
(ii) W = 289 J; Q = 0 ( insulated)
⇒ ΔU = W = 289 J
(iii) Q = 1 KJ , W = 0 ( isovolumetric process)
⇒ ΔU = Q = 1 KJ
(iv) isothermal ( constant temperature )
ΔU = Cv * ΔT∴ ΔT = 0° ( isothermal )
⇒ ΔU = 0 J
(v) isobaric ( constant pressure )
⇒ ΔU = Q + W
∴ Q = 15.6 KJ
∴ W = - ∫ P dV = - P ΔV; W (-) the system performs a job and the volume increases
.
∴ P = 950 KPa * ( 1000 Pa / KPa ) = 950000 Pa = 950000 J/m³
∴ ΔV = 18 - 5 = 13 L * ( m³ / 1000 L ) = 0.013 m³
⇒ W = - ( 950000 J/m³) * ( 0.013 m³ ) = - 12350 J ( - 12.35 KJ )
⇒ ΔU = 15.6 KJ + ( - 12.35 KJ )
⇒ ΔU = 3.25 KJ
Ответ: