One mole of a monatomic ideal gas is taken through the cycle shown in figure The pressures and temperatures at A,B etc, are denoted by P_(A),T_(A),P_(B),T_(B)etc respectively.Given T_(A)=1000K, P_(B)=(2//3)P_(A) and P_(C)=(1//3)P_(A) Then choose the incorrect option.

One mole of a monatomic ideal gas is taken through the cycle shown in

1.	One mole of a monatomic ideal gas is taken through the cycle shown in figure The pressures and temperatures at A,B etc, are denoted by P_(A),T_(A),P_(B),T_(B)etc respectively.Given T_(A)=1000K, P_(B)=(2//3)P_(A) and P_(C)=(1//3)P_(A) Then choose the incorrect option.
Answer» The work done by the gas in the process A`rarr`B is 1869.75J The heat lost by the gas sin the process B`rarr`C is -5297.25J Temperature `T_(D)` is 500K Work done from B`rarr`C is 40J Solution :For `ArarrB,T^(gamma)P^(1-gamma)=T^(gamma)P^(1-gamma)` Where `gamma=5//3` for a monatiomic gas. `T_(B)=T_(A)(P_(B))/(P_(A))^(1-(1))/(gamma) =1000(2)/(3))^(2//5)=850 K` Work dine in an adiabatic process is given by `W=(NR(T_(A)-T_(B))/(gamma-1)=(1xx8.31(1000-850))/((5//3)-1))=1869.75 J` `therefore` Process B`rarr`C is isochoric Hence W=0 and `(P_(C))/(T_(C))xT_(B)=((1//3)P_(A))/(2//3)P_(A)) T_(B)=(T_(B))/(2)=(850)/(2)=425 K` From first law of thermodynamics , `Q=DeltaU+W=nC_(V)DeltaT+0=n(3)/(2)R(T_(C)-850)` Hence `Q=1xx(3)/(2)xx8.31(125-850)=-5297.625 J` Negative sign IMPLIES that the system has lost heat. process D`rarr`A is isochoric `(P_(D))/(P_(A))=(T_(D))/(T_(A)) or P_(D)=P_(A)(T_(D))/(T_(A))` Process C`rarr`D is adiabatic Therefore, `(T_(D))/(T_(C))^(gamma)=(P_(A)T_(D))/(P_(C)T_(A))^(gamma-1)` `thereforeT^(1//gmma)=(C)[(P_(A))/(P_(C)T_(A))]^(1-1//gamma)` `rArr T^(3//5)=T_(C)[(P_(A))/((1//3)P_(A))xx1000))]^(2//5)=425 ((3)/(1000))^(2//5)` `T_(D)=500 K`