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Thermodynamics

Question
CBSEENPH11026224

1 mole of an ideal gas at an initial temperature of T K does 6R joule of work adiabatically. If the ratio of specific heats of this gas at constant pressure and at constant volume is 5/3, then final temperature of the gas will be

  • ( T - 4 ) K

  • ( T + 4 ) K

  • ( T - 2.4 ) K

  • ( T + 2.4 ) K

Solution

A.

( T - 4 ) K

In an adiabatic process, the system is insulated from the surroundings and heat absorbed or released is zero.

             PVγ  = constant

Where γ is the ratio of specific heats ( ordinary or molar ) at constant pressure and at constant volume.

Work done by the gas in adiabatic process,

          W = μ R Ti - Tfγ - 1

∴         6 R = 1 × R T - T253 - 1

⇒         6 = T - T223

⇒        T - T2 = 4

⇒         T2 = ( T - 4) K