During an experiment, an ideal gas is found to obey an additional law `VP^2=constant,` The gas is initially at a temperature T, and volume V. When it expands to a volume `2V,` the temperature becomes…….

Options

A. `(sqrt(2))T`
B. `2T`
C. `(3T)/(2)`
D. `(sqrt(3))T`

Answer

A

Explanation

According to the given problem `VT^2` = constant so the gas equation `PV = nRT` in the light of above (eliminating P) yield , `((K)/(sqrt(V)) V) = nRT`
i.e., `sqrt(V) = (nR)/(K) T`
`sqrt((V_1)/(V_2)) = ((T_1)/(T_2)) i.e., sqrt((V)/(2V)) = T/(T') or T' = (sqrt(2))T`.