A rigid diatomic ideal gas is adiabatically compressed to 1 32 of its original volume. The ratio of the final temperature to the initial temperature of the gas is:

Step-by-step solution

For a rigid diatomic ideal gas, the number of degrees of freedom is f = 5 . The ratio of specific heats is: = 1 + 2 f = 1 + 2 5 = 7 5 For an adiabatic process, the relationship between temperature and volume is: T_i V_i^ -1 = T_f V_f^ -1 Rearranging for the ratio of final to initial temperature: T_f T_i = ( V_i V_f )^ -1 We are given that V_f = V_i 32 , so V_i V_f = 32 . The exponent is - 1 = 7 5 - 1 = 2 5 . Substituting these values: T_f T_i = (32)^ 2 5 = (2^5)^ 2 5 = 2^2 = 4 Answer: 4