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Kryger [21]
3 years ago
6

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?
Chemistry
2 answers:
AnnyKZ [126]3 years ago
4 0
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
Mashcka [7]3 years ago
3 0

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.

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