1 mole of `H_(2)` gas in contained in a box of volume V 1.00 `m^(3)` at T =300 K. The gas is heated to a temperature of T= 3000K and the gas gets converted to a gas of hydrogen atoms. The final pressure would be (considering all gases to be ideal)

Options

A. same as the pressure initially
B. 2 times the pressure initially
C. 10 times the pressure initially
D. 20 times the pressure initially

Explanation

Consider the digram , when the molecules breaks into atoms the number of moles would become twice.
Now by ideal gas equation
`P= " Pressure of gas" " "n= "Number of moles"`
`R = " Gas constant" " "T = "Temperature"`
pV = nRT
As volume (V) of the container is constant.

As gases breaks number of moles becomes twice of inital so `n_(2) =2n_(1)`
`" So "" " p prop nT`
`rArr " " (P_(2))/(P_(1)) =(n_(2) T_(2))/(n_(1) T_(1))= ((2n_(1))(3000))/(n_(1)(300))=20`
`rArr " " P_(2) =200P_(1)`
Hence, final pressure of the gas would be 20 times the pressure initially.