question_answer

Starting with the same initial conditions, an ideal gas expands from volume V1 to V2 in three different ways, the work done by the gas is W2, if the process is purely isothermal, W2 if the process is purely isobaric and W3 if the process is purely adiabatic, then:

A) \[{{W}_{2}}>{{W}_{3}}>{{W}_{1}}\]        

B) \[{{W}_{1}}>{{W}_{2}}>{{W}_{3}}\]

C) \[{{W}_{3}}>{{W}_{2}}>{{W}_{1}}\]        

D) \[{{W}_{2}}>{{W}_{1}}>{{W}_{3}}\]

Correct Answer: D

Solution :

Key Idea: In a pressure-volume graph slope of adiabatic curve is maximum. The graph drawn between pressure (P) and volume (V) of a given mass of gas undergoing isothermal change is known as isothermal curve and that for a gas undergoing adiabatic change is known as adiabatic curve. The slope of adiabatic curve is maximum, the reason is that in both the isothermal and adiabatic expansion of the gas, the pressure of gas falls, but for same  fall in pressure the increase in volume of gas during adiabatic expansion is less than the increase  during  isothermal  expansion because the temperature of the gas also falls. Also slope of isobaric curve with volume axis is zero. Area under the P-V graph gives the work done by the gas. \[{{(Area)}_{2}}>{{(Area)}_{1}}>{{(Area)}_{3}}\] \[\therefore \]       \[{{W}_{2}}>{{W}_{1}}>{{W}_{3}}\]