Tuesday, May 01, 2007

Two Questions Involving Kinetic Theory of Gases

Questions based on specific heat and energy of gases often find place in all Medical and Engineering Entrance Examination question papers. See the following two typical questions:

(1) In an isobaric process, the increase in temperature of 0.4 mole of oxygen is 200K. The work done by the gas is (Universal gas constant, R = 1.99 calorie-mol –1K–1)

(a) 88 cal (b) 124.6 cal (c) 79.6 cal (d) 159.2 cal (e) 318.4 cal

Some of you may get confused on seeing the unit of work in calories. Further, R also is given in terms of calorie. Work and energy can be expressed in calorie as well as joule.

Since the pressure is constant (isobaric process), the work done by one mole of gas on getting heated through 1K is the difference (Cp – Cv) between its molar specific heats. But Cp – Cv = R, so that the work done by 0.4 mole of the gas on getting heated through 200 K is 0.4×200×R = 0.4×200×1.99 = 159.2 calorie.

(2) At what temperature will the translational kinetic energy of an ideal gas molecule be half of the value at 100º C?

(a) 323 K (b) – 179.75º C (c) 186.5º C (d) 86.5º C (e) 50º C

This is a very simple question. The translational kinetic energy of any ideal gas molecule is (3/2) kT where ‘k’ is Boltzman constant and T is the absolute (Kelvin) temperature. So, the translational kinetic energy is directly proportional to the absolute temperature. [The total kinetic energy also is directly proportional to the absolute temperature since total kinetic energy = (n/2)kT where ‘n’ is the number of degrees of freedom of the gas molecule].

Now, 100º C = 373 K. So, the energy becomes half at 186.5 K = (186.5 – 273)º C = – 86.5º C.

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