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Chemistry

A sealed tube (able to withstand a pressure of 3 atmospheres) is filled with air at 27°C and one atmosphere pressure. Find the temperature above which it will burst.

Gas Laws

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Answer

P1 = Initial pressure of the gas = 1 atm

T1 = Initial temperature of the gas = 27°C = 27 + 273 = 300 K

P2 = Final pressure of the gas = 3 atm

T2 = Final temperature of the gas = ?

V1 = V2 = V [constant volume]

By Gas Law:

P1×V1T1=P2×V2T2\dfrac{\text{P}1\times\text{V}1}{\text{T}1} = \dfrac{\text{P}2\times\text{V}2}{\text{T}2}

Substituting the values :

1×V300=3×VT2T2=3×300T2=900K\dfrac{1\times\text{V}}{300} = \dfrac{3\times\text{V}}{\text{T}2} \\[1em] \text{T}2 = 3 \times 300\\[1em] \text{T}_2 = 900 \text {K} \\[1em]

∴ The temperature above which the tube will burst is 627°C.

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