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** ^{–1}K^{–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 (C_{p} – C_{v}) between its molar specific heats. But C_{p} – C_{v} = 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) ****– 179.75****º C ****(c) ****–****186.5****º C ****(d) ****–****86.5****º C ****(e) ****50****º 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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