Question

monoatomic gas
Adiabatic Process
Pext ( =operatorname{lam}_{1} T_{1}=1 L V_{2}=2 L )
( T_{1}=T quad T_{2}=? )
( T V^{gamma-1}= ) Constant
( T_{1} V_{1}^{gamma-1}=T_{2} V_{2}^{gamma-1} )
( T_{2}=T_{1}left(frac{V_{1}}{V_{2}}right)^{gamma-1} )
( =Tleft(frac{1}{2}right)^{r-1} )
( =frac{T}{2 gamma-1}=frac{T}{2^{5} 13-1}(B) )

# One ne mole of monoatomic ideal gas expands adiabatically at initial temp. T against a constant external pressure of 1 atm from one litre to two litre. Find out the final temperature. (R = 0.0821 litre. atm K-moll) (JEE Adv. 2005] (A) T (C) T- (2) 3x0.0821 (D) T+ 0 3x0.0821

Solution