The volume of an ideal gas increases 8 times and temperature becomes (1/4)th of initial temperature during a reversible change. If there is no exchange of heat in this process ( Q = 0) then identify the gas from the following options (Assuming the gases given in the options are ideal gases) :
- NH3
- O2
- CO2
- He
Answer
(4) He
Solution
Q = 0 in a reversible change the process is adiabatic.
For an adiabatic process: TV - 1 = constant
T V - 1 = T4(8V) - 1
8 - 1 = 4
23( - 1) = 22
- 1 = 23
= 53, which is the value for a monatomic gas.
The gas is He.
Answer: (4)