JEE Main20204 Sep 2020Evening ShiftPhysicsThermodynamicsActual
The change in the magnitude of the volume of an ideal gas when a small additional pressure ∆ P is applied at a constant temperature, is the same as the change when the temperature is reduced by a small quantity ∆ T at constant pressure. The initial temperature and pressure of the gas were 300 K and 2 atm respectively. If | ΔT | = C | ΔP | then value of C in ( K / atm ) is __________
Correct answer
0
Step-by-step solution
PV = nRT PΔV + VΔP = 0 ΔV = - ΔP P V     . . . . ( i ) In second case PΔV = - nRΔT ΔV = - nRΔT P         . . . ( ii ) equating ( i ) and ( ii ) nRΔT P = - ΔP P V ΔT = ΔP   V nR c = v nR Putting the value of V ,   n and   R   ,   C   =   150