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ICSE Class 9 Chemistry Question 19 of 29

Study of Gas Laws — Question 28

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Question 28

L.P.G. cylinder can withstand a pressure of 14.9 atmosphere. The pressure gauge of the cylinder indicates 12 atmosphere at 27°C. Due to a sudden fire in the building the temperature rises. At what temperature will the cylinder explode?

Answer

P1 = Initial pressure of the gas = 12 atm

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

P2 = Final pressure of the gas = 14.9 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 :

12×V300=14.9×VT2T2=14.9×30012T2=372.5K\dfrac{12\times\text{V}}{300} = \dfrac{14.9\times\text{V}}{\text{T}_2} \\[1em] \text{T}_2 = \dfrac{14.9 \times 300}{12}\\[1em] \text{T}_2 = 372.5 \text {K} \\[1em]

∴ The temperature at which the cylinder will explode = 99.5°C.