Eg if given t p and n then v nrtp remember 1 put all

  • No School
  • AA 1
  • 58

This preview shows page 10 - 21 out of 58 pages.

e.g. if given T, P and n, then V = nRT/P Remember: (1) Put all quantities in the correct units T in K , P in atm ., V in L , n in moles. (2) Because of the simple conversion between n and m (mass) using the Molar Mass, these two quantities are virtually equivalent.
Image of page 10

Subscribe to view the full document.

Slide 11 If 2.50 grams of C 2 H 2 F 4 (g) [M=102] is placed in a 500 mL container at 10 o C, what is the pressure in torr? R = 0.082 L•atm/mol•K 1 atm. = 760 torr V = 500 mL = 0.50 L T = 10 o C = 283 K n = 2.50 g x 1 mol/102 g = 0.0245 mol 0.0245 0.082 283 0.50 L atm mol K nRT mol K P V L 1.14 atm 760 1.14 866 1 torr P atm torr atm
Image of page 11
Slide 12 19 g What is the mass of O 2 (g) in a 30 L tank at 50 o C and 400 torr? R = 0.082 L•atm/mol•K 1 atm. = 760 torr
Image of page 12

Subscribe to view the full document.

Slide 13 An Alternative Form of the Ideal Gas Equation Commonly, there is no addition or removal of material during a change in state (i.e. n 2 = n 1 ). In this case, the above relation simplifies to: Initial State (1) Final State (2) 1 1 1 1 PV R n T 2 2 2 2 PV R n T 2 2 1 1 2 2 1 1 PV PV n T n T 2 2 1 1 2 1 PV PV T T
Image of page 13
Slide 14 P V T Initial: 700 torr 5,000 in 3 40 o C = 313 K Final: 200 ?? 0 o C = 273 K Note that it is not necessary to convert volume to L or pressure to atm, but is necessary to convert temperature to K At 40 o C and 700 torr, a gas occupies a volume of 5,000 in 3 . What will the volume of the gas be at 0 o C and 200 torr? 2 2 1 1 2 1 PV PV T T 1 2 2 1 2 1 P T V V P T 3 700 273 5,000 200 313 torr K in torr K 3 2 15,300 V in
Image of page 14

Subscribe to view the full document.

Slide 15 The volume of natural gas in a storage tank at a constant pressure is 28,500 ft 3 at –15 o C (i.e. on a cold night). What volume will the gas occupy at 31 o C (i.e. on a warm day)? P V T Initial: k 28,500 ft 3 -15 o C = 258 K Final: k ?? +31 o C = 304 K Because P 2 = P 1 Note that it is not necessary to convert volume to L, but is necessary to convert temperature to K 2 2 1 1 2 1 PV PV T T 1 2 2 1 2 1 P T V V P T 2 1 1 T V T 3 3 2 2 1 1 304 28,500 33,600 258 T K V V ft ft T K
Image of page 15
Slide 16 The Density of a Gas From and Which of the following gases is the most dense at 25 o C and 0.50 atm.? A) CH 4 B) Cl 2 C) SO 2 PV nRT m n M m PV RT M m P d M V RT
Image of page 16

Subscribe to view the full document.

Slide 17 M = 120 g/mol Alternatively, one may determine the Molar Mass of a compound by measuring its gas phase density at a given P and T. A sample of an unknown gas in a 1450 mL container at 720 torr and 125 o C weighs 5.05 g. Calculate the Molar Mass of this gas. R = 0.082 L•atm/mol•K 1 atm. = 760 torr Molar Mass of a Gas from its Density m P d M V RT RT M d P mRT PV
Image of page 17
Slide 18 Gas Mixtures Mole Fraction 1 1 1 1 1 1 2 2 2 2 2 2 n 1 = moles of component 1 n 2 = moles of component 2 n tot = n 1 + n 2 = total moles 1 1 tot n X n 2 2 tot n X n
Image of page 18

Subscribe to view the full document.

Slide 19 + 1 1 1 1 1 P 1 2 2 2 2 2 P 2 1 1 2 1 1 1 2 2 2 2 P tot P tot = P 1 + P 2 X 1 = n 1 n 2 + n 1 X 2 = n 2 n 2 + n 2 Mole Fraction and Dalton’s Law - Gas mixture 1 2 tot P P P 1 1 tot P X P 2 2 tot P X P
Image of page 19
Slide 20 1.6 g H 2 [M=2] and 11.2 g CO [M=28] are placed in a 20 L tank at 32 o C Calculate: P tot , P H2 , P CO , X H2 , X CO V = 20 L & T = 32 o C = 305 K 2 1 1.6 0.80 2 H mol n g mol g 1 11.2 0.40 28 CO mol n g mol g 0.80 0.40 1.20
Image of page 20

Subscribe to view the full document.

Image of page 21
  • Fall '19

What students are saying

  • Left Quote Icon

    As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.

    Student Picture

    Kiran Temple University Fox School of Business ‘17, Course Hero Intern

  • Left Quote Icon

    I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.

    Student Picture

    Dana University of Pennsylvania ‘17, Course Hero Intern

  • Left Quote Icon

    The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time.

    Student Picture

    Jill Tulane University ‘16, Course Hero Intern

Ask Expert Tutors You can ask 0 bonus questions You can ask 0 questions (0 expire soon) You can ask 0 questions (will expire )
Answers in as fast as 15 minutes