1 liter dry air at STP expands adiabatically to a volume of 3L. If gamma=1.4, the work done by air is (3^(1.4)= 4.655) (Take air to be an ideal gas)

1 liter dry air at STP expands adiabatically to a volume of 3L. If gam

1.	1 liter dry air at STP expands adiabatically to a volume of 3L. If gamma=1.4, the work done by air is (3^(1.4)= 4.655) (Take air to be an ideal gas)
Answer» 18J 45J 90.5J 100.8J Solution :STP means `P_(1)= 1` ATM, T= 273K and `V_(1)=1L` After adiabatic expansion, `P_(2)=? And V_(2)= 3L` For adiabatic expansion `P_(1)V_(1)^(gamma) = P_(2) V_(2)^(gamma)` [ `:.T` is constant] `:. P_(2) = P_(1) ((V_(1))/(V_(2)))^(gamma)` `=1 ((1)/(3))^(gamma)` `=((1)/(3))^(1.4) [ :. gamma= 1.4]` `= (1)/(3^(1.4))= (1)/(4.655)` = 0.2148 atm WORK done in adiabatic process, `W= (P_(1)V_(1)- P_(2)V_(2))/(gamma-1)` `=(1 xx 1-0.2148 xx 3)/(1.4-1)` `=(1-0.6444)/(0.4)` `=(0.3556)/(0.4)` `W= 0.889 xx 1.01 xx 10^(5) xx 10^(-3)J` 91 atm = `1.01 xx 10^(5) (N)/(m^(2)) 1L= 10^(-3) m^(3)`] `:. W= 0.89789xx 10^(2)J` `:. W= 89.789J` `:. W= 90J`