A monatomic gas of n moles undergoes a cyclic process ABCDA as shown in figure. Process AB is isobaric, BC is adiabatic, CD is isochoric and DA is isothermal. The maximum and minimum temperature in the cycle are 4T₀ and T₀ respectively. Then

detailed solution

Correct option is A

Process DA is isothermal⇒ TD=TAFurther T∝PVFrom the graph we can see that (PV)C>(PV)D⇒ TC>TDFurther process BC is an adiabatic expansion. So, TD>TC Therefore, maximum temperature is at B and minimum temperature at D or A. Hence,TB=4T0 and TD=TA=T0In process CD:         W=0⇒ Q=ΔU                 {∵ of FLTD }Volume of the gas is constant and pressure is decreasing. Therefore, temperature and hence, internal energy will decrease ie., Q is negative or heat is released by the gas in the process CD.In process AB:Q=nCpΔT           { process is isobaric }Pressure is constant and volume is increasing. Therefore, temperature will also increase or Q is positive. Thus, heat is supplied to the gas only in process ABand Q=n5R2TB−TA⇒ Q=n5R24T0−T0=152nRT0