{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

q1r.key - Dr Zellmer Time 30 mins Chemistry 125 Winter...

Info icon This preview shows pages 1–3. Sign up to view the full content.

View Full Document Right Arrow Icon
Dr. Zellmer Chemistry 125 Thursday Time: 30 mins Winter Quarter 2012 January 26, 2012 Quiz I Name KEY Rec. TA/time 1. (6 pts) Consider a steel tank with a volume of 10.0 L that can withstand a pressure of 30.0 lb/in 2 . A sample of He gas has a volume of 50.0 L at STP. If the He gas is put in the steel tank, can liquid nitrogen (boiling point of -196°C) be used to keep the tank from exploding? (1 atm = 760 mmHg = 760 torr = 101.325 kPa = 14.7 lb/in 2 ) STP => Standard Temp. & Pressure: T = 0°C (273.15K) & P = 1 atm Since P, V & T all change use the Combined Gas Law: P 2 V 2 P 1 V 1 ------- = ------- T MUST be in kelvin T 2 T 1 Need to find T at which the pressure will be below or equal to 45.0 lb/in 2 . P 1 = 1.00 atm V 1 = 50.0 L T 1 = 0.00°C + 273.15 = 273.15 K 1 atm P 2 = 30.0 lb/in 2 × -------------- = 2.0 4 0 atm V 2 = 10.0 L T 2 = ? 14.7 lb/in 2 P 2 V 2 T 1 P 2 V 2 2.0 4 0 atm 10.0 L T 2 = ---------- = (-----)(-----)T 1 = (-------------)(---------) 273.15 K = 11 1 .489 K P 1 V 1 P 1 V 1 1.00 atm 50.0 L T 2 = 11 1 .489 K - 273.15 = & 161 .66°C = & 162°C Since N 2 boils at & 196 °C it can be used to cool the tank to & 162°C . The tank will NOT explode. Could also solve for P 2 using T 2 = -196°C and then compare P 2 to 30.0 lb/in 2 . (P 2 = 1.41 atm = 20.8 lb/in 2 ) Copyright R. J. Zellmer, January 26, 2012
Image of page 1

Info icon This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
3. (5 pts) If the temperature of a gas increases from 200.0 K to 800.0 K what happens to the average kinetic energy of the gas particles? Be specific and quantitative using Kinetic Molecular Theory to explain. Also , sketch a qualitative picture of the distribution of speeds at the two temperatures (i.e. sketch what the two curves look like relative to each other). KE avg = ½ m : 2 The average KE of a single “average” particle moving with the rms speed m = mass of the particle : = rms speed (type of average speed) KE total = 3/2 RT Total average KE per mole. Note it depends only on T and is directly proportional to T (in Kelvin). The average kinetic energy is directly proportional to the absolute temperature ( Kelvin temp.). KE avg % T (T in Kelvin) This means if the temperature increases then the average KE will increase. More specifically, if the temperature quadruples (200.0 K to 800.0 K) the KE avg quadruples (and vice versa), they are directly proportional. For 1 mole of a gas the average total KE per mole is equal to (3/2)RT.
Image of page 2
Image of page 3
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

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