Stuck with physics gas questions

• WoAiMeiMei

  10 Apr 08, 20:42

  Originally posted by eagle:

  C19 P38

  temperature the same, kinetic energy of molecules should be the same

  thus, (0.5)(m1)(v1) = (0.5)(m2)(v2)

  but v2 = 2v1,

  m1/m2 = v2/v1 = 2

   

  Not confirmed though....

  ahh i have a hint:x

   

  You do not need all the information given. Did you write the equations for average speed and rms speed? How is molar mass related to molecular mass?

• eagle

  10 Apr 08, 21:10

  that's O level chemistry I think...

  molar mass = molecular mass * mole number

• eagle

  10 Apr 08, 21:13

  Originally posted by eagle:

  C19 P38

  temperature the same, kinetic energy of molecules should be the same

  thus, (0.5)(m1)(v1) = (0.5)(m2)(v2)

  but v2 = 2v1,

  m1/m2 = v2/v1 = 2

   

  Not confirmed though....

  Oops... Careless with the above formula... Should be 1/2 m v^2

  (0.5)(m1)(v1)^2 = (0.5)(m2)(v2)^2

  but v2 = 2v1,

  m1/m2 = (v2/v1)^2 = 4

• WoAiMeiMei

  10 Apr 08, 21:25

  i think that's related to the value "R"

   

  ps: 4 isnt the ans

• eagle

  10 Apr 08, 21:32

  With this formula
  p = (nMv^2)/3V

  pV/n = Mv^2 / 3

  this is a constant because R and T are constants here

  so,

  m1v1^2 / 3 = m2v2^2 / 3

  Ratio of m1/m2 is still 4 leh...

   

  I think different p means different V or n only... The Vrms is still heavily dependent on temperature.

• WoAiMeiMei

  10 Apr 08, 21:46

  arrr....

  use Vavg and Vrms equations and rearrange to get m1/m2 ratio

  thats how someone else got it >__< but how :O

• eagle

  10 Apr 08, 21:49

  OH!!!!!! SORRY.... didn't see it is Vavg!

  According to your equations,

  Vrms2 = Vavg * sqrt (3 * pi / 8) = 2.171 Vrms1

  So should be 2.171^2 = 4.71?

   

• WoAiMeiMei

  10 Apr 08, 22:05

  4.71 is the correct answer! XD

• eagle

  10 Apr 08, 22:40

  apologize for being so careless :(

• eagle

  10 Apr 08, 22:54

  C19 P43

  a) 0J as volume didn't change

  b)

  change in T = 16.3
  N = 2.42 mol
  f = degree of freedom = 3 for monoatomic gas
  k = boltzmann constant

  change in thermal energy = 491.7 J => not confirmed

  c)

  change in internal energy = a) + b) = 491.7J

  d) should be 491.7J / (2.42 * 1mol) = 3.37e-22 J

• WoAiMeiMei

  11 Apr 08, 07:08

  thanks, that's the correct answer ^^

• eagle

  11 Apr 08, 11:44

  C19 P61

  a)

  1.8 * 176^y = 5.51 * 79.2^y
  2.2222^y = 3.061111
  y = ln(3.061111) / ln(2.222222) = 1.4

  => diatomic gas

  b)

  so,
  310 * 176^1.4 = T * 79.2^1.4
  T = 948.1K

  c) PV = nRT
  so n = PV/RT
  p = 1.80atm, 1 atm = 101325 Pa
  V = 176L = 176000cm3 = 0.176 m3
  R = 8.314 J·K⁻¹mol^-1
  T = 310K

  thus, n = 6.92 mol

• WoAiMeiMei

  11 Apr 08, 15:15

  b and c cant get correct...hint is to apply the ideal gas law. anyways cant reply tonight.. 

• eagle

  11 Apr 08, 18:50

  Originally posted by eagle:

  C19 P61

  a)

  1.8 * 176^y = 5.51 * 79.2^y
  2.2222^y = 3.061111
  y = ln(3.061111) / ln(2.222222) = 1.4

  => diatomic gas

  b)

  so,
  310 * 176^1.4 = T * 79.2^1.4
  T = 948.1K

  c) PV = nRT
  so n = PV/RT
  p = 1.80atm, 1 atm = 101325 Pa
  V = 176L = 176000cm3 = 0.176 m3
  R = 8.314 J·K⁻¹mol^-1
  T = 310K

  thus, n = 6.92 mol

  Part b careless.... Equation is still correct... Anyway you should also help check for careless mistake mah... Not just take the values and use them without trying to understand how the equation was used...

  The power should be 0.4, not 1.4

  310 * 176^0.4 = T * 79.2^0.4
  T = 426.65K

   

  Part c is already ideal gas equation. Maybe you could help check if the calculations are wrong...

• WoAiMeiMei

  12 Apr 08, 10:46

  hmm im not sure for part c... i keep hitting a wall... will redo tonight.. still soo tired from last night

• FirePig

  12 Apr 08, 16:33