Advertisement

Upcoming SlideShare

Help! How to find A and B- Du Point Analysis- Calculate and interpret.docx

Loading in ... 3

Jan. 27, 2023•0 likes## 0 likes

•3 views## views

Be the first to like this

Show More

Total views

0

On Slideshare

0

From embeds

0

Number of embeds

0

Download to read offline

Report

Education

Helium gas with a volume of 2.90 L under a pressure of 1.80 atm and at a temperature of 43.0 ?C is warmed until both the pressure and volume of the gas are doubled. What is the final temperature? How many grams of helium are there? The molar mass of helium is 4.00 g/mol Solution a) Using the gas equation, P1V1/T1 = P2V2/T2 Here let P1 = 1.8 atm so P2 = 2P1 atm V1 = 2.90 L so V2 = 2V1 L T1 = 273 + 43 = 316 K , T2 = ? T2 = P2V2/P1V1*T1 = 2P1*2V1/(P1V1)*316 = 1264 K b) By using, PV = nRT where n is the No. of moles of He n = PV/RT = 1.8*2.90/(0.0821*316) = 0.20 Thus the moles of He = 0.20 And the mass of He = 0.20 * 4 = 0.80 g .

rtodd588Follow

Advertisement

Advertisement

Advertisement

Help! How to find A and B- Du Point Analysis- Calculate and interpret.docxrtodd588

Help with section D- (Second picture) finding the T(s) and f (Hz)Solut.docxrtodd588

Help story Bookmarks Window saplinglearning-com Hofstra U Sapling Lear.docxrtodd588

help please )- True or False and explain why- Aluminum-silicate result.docxrtodd588

Hello- please let me know the MR- Thank you- Q-10-0-1p Q Maginal Reven.docxrtodd588

help please Write force F in Cartesian vector form What is the moment.docxrtodd588

- Helium gas with a volume of 2.90 L under a pressure of 1.80 atm and at a temperature of 43.0 ?C is warmed until both the pressure and volume of the gas are doubled. What is the final temperature? How many grams of helium are there? The molar mass of helium is 4.00 g/mol Solution a) Using the gas equation, P1V1/T1 = P2V2/T2 Here let P1 = 1.8 atm so P2 = 2P1 atm V1 = 2.90 L so V2 = 2V1 L T1 = 273 + 43 = 316 K , T2 = ? T2 = P2V2/P1V1*T1 = 2P1*2V1/(P1V1)*316 = 1264 K b) By using, PV = nRT where n is the No. of moles of He n = PV/RT = 1.8*2.90/(0.0821*316) = 0.20 Thus the moles of He = 0.20
- And the mass of He = 0.20 * 4 = 0.80 g

Advertisement