A certain volume of a gas at 300 K expands adiabatically until its volume is doubled. The resultant fall in temperature of the gas is nearly (The ratio of the specific heats of the gas =1.5 )

Step-by-step solution

For adiabatic process, aligned & V₂=2 V₁, =1.5, T₁=300 ~K & TV ^ -1 = constant T ₁ ~V ₁^ -1 = T ₂ ~V ₂^ -1 & 300 ( ~V ₁ )^ 0.5 = T ₂ (2 ~V ₁ )^ 0.5 & ~T ₂=300 ( 1 2 )^ 0.5 aligned Fall in temperature is T = T ₁- T ₂=300 (1- 1 2 )=88 ~K