What experimental factors are assumed to be constant in this experiment? DATA AN
ID: 560065 • Letter: W
Question
What experimental factors are assumed to be constant in this experiment?
DATA AND CALCULATIONS Volume (mL) Pressure (kPa) Constant, k (P/V or P.V) mu ollox 190.0 K mt 90 K 19.do nf 51.84 kfa 630 PROCESSING THE DATA 1. If the volume is doubled from 5.0 mL to 10.0 mL, what does your data show happens to the ressure? Show the pressure values in your answer. If the Volume of 5.0m. rpad o-tn Shouss that the pressure itself csovd &-EbUcE dotud 2. If the volume is halved from 20.0 mL to 10.0 mL, what does your data show happens to the 2rn 15t 9 00 d'oee an inereas'e of pressure fsica ihe original laiuoe 3. If the volume is tripled from 5.0 mL to 15.0 mL, what does your data show happened to the ressure? Show the pressure values in your answer. the volume is haivedl from 80.om ressure? Show the pressure values in your answer.f voorna isnNa From OrL original Pressure vaue wilt decree by its original value. S.mL, ten sols-3ito.oo From your answers to the first three questions and the shape of the curve in the plot of pressure versus volume, do you think the relationship between the pressure and volume of a confined gas is direct or inverse? Explain your answer. Thod nversa sina Nec. cFemains thi Samu. whireas, volume would mversay as porions wi-th 4. resSure.if there wnsan increase with Pressore there woid nied to be o 5. Based on your data, what would you expect the pressure to be if the volume of the syringe was increased to 40.0 mL? Explain or show work to support your answer. Based on your data, what would you expect the pressure to be if the volume of the syrin was decreased to 2.5 mL? Explain or show work to support your answer. 6.Explanation / Answer
The experiment assumes that the temperature and mass of the gas are kept fixed.
5) We have P1 = 101.60 kPa, V1 = 10.0 mL and V2 = 40.0 mL; use Boyle’s law
P1*V1 = P2*V2
====> (101.60 kPa)*(10.0 mL) = P2*(40.0 mL)
====> P2 = (101.60 kPa)*(10.0 mL)/(40.0 mL) = 25.40 kPa
The pressure falls to 25.40 kPa (ans).
6) We have P1 = 190.00 kPa, V1 = 5.0 mL and V2 = 2.5 mL; use Boyle’s law
P1*V1 = P2*V2
====> (190.00 kPa)*(5.0 mL) = P2*(2.5 mL)
====> P2 = (190.00 kPa)*(5.0 mL)/(2.5 mL) = 380.00 kPa
The pressure increases to 380.00 kPa (ans).
Related Questions
drjack9650@gmail.com
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.