A sealed tube which can withstand a pressure of 3 atmosphere of 3 atmosphere is filled with air at ${27}^0$C and 760 mm pressure. Find the temperature above which it will burst.

The correct option is A

627⁰C

Let the volume of air in the tube be V ml.

Given conditions New conditions

$V_1$ = V ml $V_2$ = V ml

$P_1$ = 760 mm = 1 atm $P_2$ = 3 atm

$T_1$ = 273+27 = 300K $T_2$ = ?

By applying gas equation, we have

$\frac{V\times 1}{300}=\frac{V\times 3}{T_2}$

$T_2=300\times 3$ = 900K

Thus the temperature above which the tube will burst = 900 – 273 = ${627}^0$C.