Question

A 400 mL sample of nitrogen in a sealed, inflexible container has a pressure of 1200 torr at a temperature of 250 K. It is known that the container will rupture at a pressure of 1800 torr. At what temperature will the container rupture?

Answer 1

by  use     of   gay   lussacs   law
state  b  that  for  a  given  mass  and  constant  volume  of   an  ideal   gas  .the pressure  exerted  on    the  side   of  its   container  is  directly  proportional   to  its  obsolute  temperature
T2=  (p2  x  T1)/P1
  
(1800  x250)  /  1200=  375K

Answer 2

Answer:

Container will rupture at temperature of 375 K.

Explanation:

Initial pressure of the  nitrogen gas =$P_1= 1200 torr = 1.572 atm$

(1 torr = 0.00131 atm)

Initial temperature of nitrogen gas =$T_1= 250 K$

Final pressure of the nitrogen gas =$P_2=1800 torr=2.358 atm$

Final temperature of nitrogen gas =$T_2=?$

Since, the container is inflexible that is volume remains constant we can apply Gay Lussac's law:

$\frac{P_1}{T_1}=\frac{P_2}{T_2}$

$\frac{1.572 atm}{250 K}=\frac{2.358 atm}{T_2}$

$T_2=375 K$

Container will rupture at temperature of 375 K.