As the day advances, the temperature of an open room of volume `41. 6m^3` increases from `26.85^@C " to " 28^@C` . The atmospheric pressure in the room remains `1xx 10^5` Pa. Then difference in the number of molecules present in the room after and before the heating is `n xx 10^(24)` (Take `1K = 273. 15^@C` Universal gas constant `R= 8.32 "Jmol"^(-1) K^(-1)` , Avogadro's number `N_A = 6.023 xx 10^(23)` ). The value of n is

Answer

`-3.833`

Explanation

Using ideal gas equation,
Before heating, at
`T_1 = 26.85 + 273. 15 = 300K`
` PV = n_1 R xx 300 .... (i)`
After heating, at
`T_2 = 28 + 273. 15 = 301.15K,`
` PV = n_2R xx 301. 15 .... (ii)`
where, `n_1 and n_2` are number of moles at `T_1 and T_2` respectively.
From equations (i) and (ii),
`n_2-n_1=(PV)/(Rxx301.15) - (PV)/(R xx 300)`
But, number of molecules `= n xx N_A`
`:. n_f - n_i = (n_2 – n_1)N_A`
`:. n_f -n_i = (-PV)/R xx ((1.15)/(301.15 xx300))xx 6.023xx10^(23)`
Given : `P = 10^5` P a and `V = 41.6 m63`
`:.` Number of molecules
`n_f -n_i= (-10^5 xx4.16 xx1.15 xx 6.023 xx10^(23))/(8.32 xx301.15 xx 300)`
`=-(10^5 xx5 xx1.15 xx10^(23))/(50xx300)`
`=- (1.15 xx10^(27))/(300)`
`:. n_f -n_i = -3.833 xx10^(24)`