{"id":2296,"date":"2018-04-30T07:44:10","date_gmt":"2018-04-29T23:44:10","guid":{"rendered":"https:\/\/gurumuda.net\/physics\/?p=2296"},"modified":"2023-08-08T13:07:58","modified_gmt":"2023-08-08T13:07:58","slug":"kinetic-theory-of-gas-and-first-law-of-thermodynamics-problems-and-solutions","status":"publish","type":"post","link":"https:\/\/gurumuda.net\/physics\/kinetic-theory-of-gas-and-first-law-of-thermodynamics-problems-and-solutions.htm","title":{"rendered":"Kinetic theory of gas and first law of thermodynamics \u2013 problems and solutions","gt_translate_keys":[{"key":"rendered","format":"text"}]},"content":{"rendered":"

Kinetic theory of gas and first law of thermodynamics \u2013 problems and solutions<\/span><\/p>\n

1. Ideal gases are in a container with a volume of 4 liters and its pressure is 3 atm (1 atm = 105<\/sup> N.m-2<\/sup>). The ideal gases<\/a> heated at a constant pressure from 27o<\/sup>C to 87o<\/sup>C. The heat capacity<\/a> of the gas is 9 J.K-1<\/sup>. What is the final volume of gases and the change of internal energy of gases?<\/span><\/p>\n

Solution<\/span><\/p>\n

Isobaric process (constant pressure)<\/b><\/span><\/p>\n

Known :<\/u><\/span><\/p>\n

The initial volume of gas (V1<\/sub>) = 4 liters<\/span><\/p>\n

The initial temperature<\/a> of gas (T1<\/sub>) = 27o<\/sup>C + 273 = 300 K <\/span><\/p>\n

The final temperature of gas (T2<\/sub>) = 87o<\/sup>C + 273 = 360 K <\/span><\/p>\n

The pressure of gas (P) = 3 atm = 3 x 105<\/sup> N.m-2<\/sup><\/span><\/p>\n

The heat capacity of gas (C) = 9 J.K-1<\/sup><\/span><\/p>\n

Wanted :<\/u> The final volume of gas (V2<\/sub>) and the change of internal energy of gas (\u0394U)<\/span><\/p>\n

Solution :<\/u><\/span><\/p>\n

Determine the final volume using the equation of Charles’s law<\/a> (isobaric process or constant pressure) :<\/span><\/p>\n

\"Kinetic<\/span><\/p>\n

The change in volume :<\/u><\/span><\/p>\n

1 liter = 0.001 m3 <\/sup><\/span><\/p>\n

Initial volume (V1<\/sub>) = 4 (0.001 m3<\/sup>) = 0.004 m3<\/sup><\/span><\/p>\n

Final volume (V2<\/sub>) = 4.8 (0.001 m3<\/sup>) = 0.0048 m3<\/sup><\/span><\/p>\n

The change in volume (\u0394V) = V2<\/sub> \u2013 V1 <\/sub>= 0.0048 m3<\/sup> \u2013 0.004 m3<\/sup> = 0.008 m3<\/sup>. <\/span><\/p>\n

The change in temperature :<\/u><\/span><\/p>\n

The change in temperature (\u0394T) = T2<\/sub> \u2013 T1<\/sub> = 360 K – 300 K = 60 K <\/span><\/p>\n

Determine the change of the internal energy (\u0394U) of the ideal gas using the equation of the first law of thermodynamics.<\/span><\/p>\n

\u0394U = Q \u2013 W<\/span><\/p>\n

\u0394U = the change of the internal energy, Q = heat, W = work<\/i><\/span><\/p>\n

Determine work (W) at constant pressure :<\/span><\/p>\n

W = P \u0394V = (3 x 105<\/sup>)(0.0008) = (3 x 101<\/sup>)(8) = (30)(8) = 240 Joule <\/span><\/p>\n

Determine heat (Q) using the equation of the heat capacity (C) :<\/span><\/p>\n

C = Q \/ \u0394T<\/span><\/p>\n

Q = (C)(\u0394T) = (9)(60) = 540 Joule<\/span><\/p>\n

Determine the change of the internal energy :<\/span><\/p>\n

\u0394U = Q \u2013 W = 540 Joule – 240 Joule = 300 Joule.<\/span><\/p>\n

2. 6 liters of ideal gases at 2 atm are in a container (1 atm = 105<\/sup> N.m-2<\/sup>). The gas heated from 27o<\/sup>C to 77o<\/sup>C at a constant pressure. If the heat capacity of gas is 5 J.K-1<\/sup>, what is the final volume and the change of internal energy of the gas.<\/span><\/p>\n

Solution :<\/span><\/p>\n

Isobaric process<\/a> (constant pressure)<\/b><\/span><\/p>\n

Known :<\/u><\/span><\/p>\n

The initial volume of the ideal gases (V1<\/sub>) = 6 liters<\/span><\/p>\n

The initial temperature of the ideal gases (T1<\/sub>) = 27o<\/sup>C + 273 = 300 K <\/span><\/p>\n

The final temperature of the ideal gases (T2<\/sub>) = 77o<\/sup>C + 273 = 350 K <\/span><\/p>\n

The pressure of the ideal gases (P) = 2 atm = 2 x 105<\/sup> N.m-2<\/sup><\/span><\/p>\n

The heat capacity of gases (C) = 5 J.K-1<\/sup><\/span><\/p>\n

Wanted:<\/u> The final volume of the gas (V2<\/sub>) and the change of internal energy of the gas (\u0394U)<\/span><\/p>\n

Solution :<\/u><\/span><\/p>\n

Determine the final volume of gas using the equation of Charles’s law (isobaric process or constant pressure) :<\/span><\/p>\n

\"Kinetic<\/span><\/p>\n

The change in volume :<\/u><\/span><\/p>\n

1 liter = 0.001 m3 <\/sup><\/span><\/p>\n

The initial volume (V1<\/sub>) = 6 (0.001 m3<\/sup>) = 0.006 m3<\/sup><\/span><\/p>\n

The final volume (V2<\/sub>) = 7 (0.001 m3<\/sup>) = 0.007 m3<\/sup><\/span><\/p>\n

The change in volume (\u0394V) = V2<\/sub> \u2013 V1<\/sub> = 0.007 m3<\/sup> \u2013 0.006 m3<\/sup> = 0.001 m3<\/sup><\/span><\/p>\n

The change in temperature :<\/u><\/span><\/p>\n

The change in temperature (\u0394T) = T2<\/sub> \u2013 T1 <\/sub>= 350 K – 300 K = 50 K <\/span><\/p>\n

Determine the change of the internal energy (\u0394U) of the ideal gases using the equation of the first law of thermodynamics.<\/span><\/p>\n

\u0394U = Q \u2013 W<\/span><\/p>\n

\u0394U = the change in the internal energy, Q = heat, W = work<\/i><\/span><\/p>\n

Determine work (W) at constant pressure :<\/span><\/p>\n

W = P \u0394V = (2 x 105<\/sup>)(0.001) = (2 x 102<\/sup>)(1) = (200)(1) = 200 Joule <\/span><\/p>\n

Determine heat (Q) using the equation of the heat capacity (C) :<\/span><\/p>\n

C = Q \/ \u0394T<\/span><\/p>\n

Q = (C)(\u0394T) = (5)(50) = 250 Joule<\/span><\/p>\n

Determine the change in the internal energy :<\/span><\/p>\n

\u0394U = Q \u2013 W = 250 Joule – 200 Joule = 50 Joule.<\/span><\/p>\n

    \n
  1. What are the main assumptions of the kinetic theory of gases?<\/strong><\/span>\n