A monatomic gas at 630 K expands adiabatically to 27 times its initial volume. The final temperature of the gas is

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Correct answer: D. (d) For mono-atomic, f=3 =1+ 2f= 53 For adiabatic PV^ = constant ( ∴ PV= nRT ) ( nRTV )V^ = constant T_1 V_1^ -1=T_2 V_2^ -1 ⇒ 630 ( V )^ -1 (27 V )^ -1=T_2 Final temperature of gas, T_2= 630 (27 )^ -1 = 630 (3 )^3 53-1=70 K

$A monatomic gas at 630 K expands adiabatically to 27 times its initial volume. The final temperature of the gas is$

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