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-

P_1T_1= P_2T_2 #160; P_1= 12 atm, T_1= 27^oC= 300K, P_2=14.9atm ⇒ T_2= 14.9× 30012=372.5K = 99.5^oC #160; ...

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Standard XII

Chemistry

Avogadro's Law & Gay Lussac's Law

L.P.G. cylind...

Question

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

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Solution

$\frac{P_{1}}{T_{1}} = \frac{P_{2}}{T_{2}}$
$P_{1} = 12 a t m, T_{1} = 27^{o} C = 300 K, P_{2} = 14.9 a t m$

$\Rightarrow T_{2} = \frac{14.9 \times 300}{12} = 372.5 K = {99.5}^{o} C$

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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?

P1T1=P2T2 P1=12 atm,T1=27oC=300K,P2=14.9atm⇒T2=14.9×30012=372.5K=99.5oC

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

$\frac{P_{1}}{T_{1}} = \frac{P_{2}}{T_{2}}$
$P_{1} = 12 a t m, T_{1} = 27^{o} C = 300 K, P_{2} = 14.9 a t m$

$\Rightarrow T_{2} = \frac{14.9 \times 300}{12} = 372.5 K = {99.5}^{o} C$