A spherial balloon contains air at temperature `T_(0)` and pressure `P_(0)`. The balloon material is such that the instantaneous pressure inside is proportional to the square of the diameter. When the volume of the balloon doubles as a result of heat transfer, the expansion follows the law

Options

A. `PV =` constant
B. `PV^(2//5) =` constant
C. `PV^(-1) =` constant
D. `PV^(-2//3) =` constant

Answer

D

Explanation

d. Pressure is directly proportional to square of the diameter of the balloon,
`(P_(2))/(P_(1)) = (D_(2)^(2))/(D_(1)^(2))`
When `V_(2) = 2 V_(1) implies D_(2)^(3) = 2 D_(1)^(3)`
`implies (D_(2))/(D_(1)) = ((V_(2))/(V_(1)))^(1//3)`
Form Eqs. (i) and (ii),
`(P_(2))/(P_(1)) = ((V_(2))/(V_(1)))^(2//3)`
`PV^(-2//3) =` constant