0 . 02 moles of an ideal diatomic gas with initial temperature 20 ° C is compressed from 1500   cm 3 to 500   cm 3 . The thermodynamic process is such that p V 2 = β , where β is a constant. Then, the value of β is close to (the gas constant, R = 831   J / K / mol ).

Step-by-step solution

Process equation is p V 2 = β . . . ( i ) As gas is ideal, it obeys gas equation, p V = n R T . . . ( iii ) From Eqs. ( i ) and ( ii ) , gives ( n R T ) · V = β Here, n = 0 . 02 moles, R = 8 . 31   JK - 1   mol - 1 ,   T = 20 ° C + 273 = 293   K and V = 1500   cm 3 = 1 . 5 × 10 - 3   m 3 ∴   β = 0 . 02 × 8 . 31 × 293 × 1 . 5 × 10 - 3 = 7 . 3 × 10 - 2   Pa - m 6 ≈ 7 . 5 × 10 - 2   Pa - m 6