JEE Main 2019 Physics question
A rigid diatomic ideal gas undergoes an adiabatic process at room temperature. The relation between temperature and volume for this process is = constant, then x is:
Answer: (B)
Solution
x = γ − 1 = 7/5 − 1 = 2/5.
A rigid diatomic ideal gas undergoes an adiabatic process at room temperature. The relation between temperature and volume for this process is = constant, then x is:
x = γ − 1 = 7/5 − 1 = 2/5.