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-

=12 ATM or 273+27=300 K =14.9 ATM therefore, = =372.5 K or 372.5 273= ...

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Standard X

Chemistry

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

=12 ATM

or 273+27=300 K

=14.9 ATM

therefore, $T subscript 2 equals fraction numerator P subscript 2 T subscript 1 over denominator P subscript 1 end fraction$

= $fraction numerator 14.9 cross times 300 over denominator 12 end fraction$

=372.5 K or 372.5 -273=

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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.

=12 ATM or 273+27=300 K =14.9 ATM therefore, = =372.5 K or 372.5 -273=

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.

=12 ATM

or 273+27=300 K

=14.9 ATM

therefore, $T subscript 2 equals fraction numerator P subscript 2 T subscript 1 over denominator P subscript 1 end fraction$

= $fraction numerator 14.9 cross times 300 over denominator 12 end fraction$

=372.5 K or 372.5 -273=