The pressure P 1 and density d 1 of diatomic gas γ = 7 5 changes suddenly to P 2 > P 1 and d 2 respectively during an adiabatic process. The temperature of the gas increases and becomes____times of its initial temperature. (given d 2 d 1 = 32 )

Step-by-step solution

For adiabatic process, P V γ = constant We know, density d = mass Volume . Then, we have P d γ = constant Given here, d 2   d 1 = 32 and γ = 7 5 . The ratio P 1 P 2 = d 1   d 2 γ = 1 32 7 5 = 1 128 The ratio of temperature using ideal gas equation is T 1   T 2 = P 1   V 1 P 2   V 2 = 1 128 × 32 = 1 4 Thus, T 2 = 4 T 1 .