Question

A sealed tube which can withstand
a pressure of 3 atm is filled with air
at 27°C 1 atm pressure. Find the
temperature above which it will burst​

Answer 1

Answer:

$The \: temperature \: with \: which \\ \: the \: tube \: will \: burst \: \: be \: = > \\ \: 900 \: K \: (or \: 627 \: °C= 900 - 273)$

Explanation:

$Let \: the \: volume \: of \: air \: in \: tube \: be \\ \: V \: \: cm {}^{3} \\ that \: is, \\ Initial \: volume \: \: V _{1} \: = V \: \: cm {}^{3} \\ Final \: volume \: \: V _{2} \: = V \: \: cm {}^{3} \\ Initial \: pressure \: \: P _{1} = 1 \: atm \\ Final \: pressure \: \: P _{2} \: = 3 \: atm$

$Initial \: temperature \: \: \\ T_{1} = 273 + 27 = 300 \: K \\ Final \: temperature \: \: T _{2} \: = \: ?$

$According \: to \: the \: gas \: equation \\ \frac{P _{1}V_{1}}{T _{1}} = \frac{P _{2}V _{2} }{T _{2} } \\ = > \frac{1 \: atm \times V \: cm {}^{3} }{300 \: K} = \frac{3 \: atm \times V \: cm {}^{3} }{T _{2} } \\ Therefore \: \: T _{2} = \frac{300 \: K \times 3 \: atm \times V \: cm {}^{3} }{1 \: atm \times V \: cm {}^{3} } \\ T _{2} = 300 \: K \times 3 \\ Final \: temperature = > 900 \: K$