One mole of an ideal diatomic gas expands from volume V to 2V isothermally at a temperature 27 and does W joule of work. If the gas undergoes same magnitude of expansion adiabatically from 27 doing the same amount of work W, then its final temperature will be (close to) . ( e 2 = 0.693)
- -56
- -117
- -30
- -189
Answer
(1) -56
Solution
Isothermal work: W = nRT e V2V1 = 1 R 300 0.693 = 207.9R
Adiabatic work (diatomic, = 75): W = nR(T1 - T2) - 1 = 52R(T1 - T2)
Equating: 52R(T1 - T2) = 207.9R
T1 - T2 = 2 207.95 = 83.2 K
T2 = 300 - 83.2 = 216.8 K
t2 = 216.8 - 273 -56
Answer: (1)