One mole of an ideal diatomic gas expands from volume V to 2 V isothermally at a temperature 27^ C and does W joule of work. If the gas undergoes same magnitude of expansion adiabatically from 27^ C doing the same amount of work W , then its final temperature will be (close to) _ _ _ _ ^ C . ( _ e 2=0.693 )

Step-by-step solution

For isothermal expansion of 1 mole diatomic gas from V to 2V at 300 K: W = RT 2 = R(300) 2 For adiabatic expansion with same volume change and same work output: W = nC_V T = C_V(T_i - T_f) For diatomic gas, C_V = 5 2 R . Setting work equal: R(300) 2 = 5 2 R(300 - T_f) Simplifying: 300 2 = 2.5(300 - T_f) , so 300(0.693) = 2.5(300 - T_f) 207.9 = 750 - 2.5T_f , giving T_f = 216.84 K -56°C.