Question

A box contains `N` molecules of a perfect gas at temperature `T_(1)` and temperature `P_(1)`. The number of molecule in the box is double keeping the total kinetic energy of the gas same as before. If the new pressure is `P_(2)` and temperature `T_(2), then
A. `P_(2) = P_(1), T_(2) = T_(1)`
B. `P_(2) = P_(1), T_(2) = (T_(1))/(2)`
C. `P_(2) = 2P_(1), T_(2) = T_(1)`
D. `P_(2) = 2P_(1), T_(2) = (T_(1))/(2)`

Answer 1

Correct Answer - B
b. Kinetic energy of `N` molecules of gas,
`E = (3)/(2) N kT`
Initially
`E_(1) = (3)/(2) N_(1) kT_(1)`
and finally
`E_(2) = (3)/(2) N_(2) kT_(2)`
But according to the problem `E_(1) = E_(2)` and `N_(2) = 2N_(1)`
`(3)/(2) N_(1) kT_(1) = (3)/(2) (2N_(1)) kT_(2)`
`implies T_(2) = (T_(1))/(2)`
Since the kinetic energy is constant
`(3)/(2) N_(1) kT_(1) = (3)/(2) N_(2) kT_(2)`
`implies N_(1) T_(2) = N_(2) T_(2)`
`:. NT =` constant
From ideal gas equation of `N` molecules `PV = NkT`
`implies P_(1) V_(1) = P_(2) V_(2)`
`:. P_(1) = P_(2)`
(as `V_(1) = V_(2)` and `NT =` constant)