A sample of 1 mole gas at temperature T is adiabatically expanded to double its volume. If adiabatic constant for the gas is = 32, then the work done by the gas in the process is :
- RT[2 + 2]
- RT[2 - 2]
- RT[2 - 2]
- TR[2 + 2]
Answer
(2) RT[2 - 2]
Solution
Adiabatic: TV - 1 = constant
T2 = T ( V2V )1/2 = T 2
Work done by the gas: W = nR(T1 - T2) - 1 = R (T - T 2 )1/2
= 2RT (1 - 1 2 ) = RT[2 - 2]
Answer: (2)