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
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 =
∴ 6 R =
⇒ 6 =
⇒ T - T2 = 4
⇒ T2 = ( T - 4) K