At what temperature the volume of 28 grams of `N_2` will be 1L exerting a pressure of 1 atm?

Explanation

28g of `N_2` at 273K, 1 atm pressure occupies 22.4L
Charles. law is given as,` (V_1)/(V_2) = (T_1)/(T_2)`
`V_1 = 22.4 L " " T_1 = 273 K`
` V_2 = 1.0 L " " T_2 =?`
The temperature at which 1L of 28g of N, exerts a pressure of 1 atm =` T_2 = (V_2T_1)/(V_1) = ( 1 xx 273 )/( 22.4 ) = 12.2 K =- 260.8^@ C `