Question

`28 g "of" N_2` gas is contained in a flask at a pressure of `10 atm` and at a temperature of `57^@ C`. It is found that due to leakage in the flask, the pressure is reduced to half and the temperature to `27^@ C`. The quantity of `N_(2)` gas that leaked out is.
A. `11//20 g`
B. `20//11 g`
C. `5//63 g`
D. `63//5 g`

Answer 1

Correct Answer - D
`28 g of N_2 mean 1 g - mol`
Now, `n = (p V)/(R T) or n prop (P)/(T)`
`(n_f)/(n_i)= ((p_f)/(p_i))((T_i)/ (T_f))= (1/2) ((273 + 57)/(273 +27))= (11)/(20)`
or `n_f = ((11)/(20))n_i = (11)/(20) mol`
`Delta n = n_i - n_f = 1- (11)/(20) = (9)/(20) g - mol`
`Delta m = (9)/(20) xx 28 g = (63)/(5) g - mol`.