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

Container will rupture at temperature of 375 K.

Step-by-step 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:

`(P_1)/(T_1)=(P_2)/(T_2)`

`(1.572 atm)/(250 K)=(2.358 atm)/(T_2)`

`T_2=375 K`

Container will rupture at temperature of 375 K.

Answer 2

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