Correct Answer - B
Here, `V_(1) = 30m^(3), T_(1) = 17+273 = 290 K`
`P_(1) = 1 xx 10^(5)Pa`
`V_(2) = 30 m^(3) , T_(2) = 27+273 = 300K`,
`P_(2) = 1 xx 10^(5) Pa`.
Let `N_(1),N_(2)` be the no, of moles of a gas at temperature `T_(1) and T_(2)` respectively. then
`N_(1) =(P_(1)V_(1))/(RT_(1)) = ((1 xx 10^(5)) xx 30)/(83 xx 290) = 1.24 xx 10^(3)`
or `N_(2) = (P_(2)V_(2))/(RT_(2)) = ((1 xx 10^(5)) xx 30)/(83 xx 300) = 1.20 xx 10^(3)` ltbr. Change in the number of moles
`N_(2)-N_(1) =(1.20-1.24) xx 10^(3) = -0.04 xx 10^(3)`
Change in the number of molecules
`n_(f)-n_(i) = (N_(2)-N_(1)) xx (6.023 xx 10^(23))`
`= - (0.04 xx 10^(3)) xx (6.023 xx 10^(23))`
=-2.5 xx 10^(25)`.