LPG cylinder can withstand a pressure of 14.9 atm. The pressure gauge of the cylinder reads 12 atm at 27^°C. Due to a sudden fire in the building the temperature rises. Are what temperature will the cylinder explode?

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Solution

P1=12 ATM

T1=27'C or 273+27=300 K

P2=14.9 ATM

therefore,
P1/T1 = P2/T2
=>T2=(P2 x T1)/P1

=(14.9 x 300)/12

=372.5 K or 372.5 -273=99.5'C

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LPG cylinder can withstand a pressure of 14.9 atm. The pressure gauge of the cylinder reads 12 atm at 27°C. Due to a sudden fire in the building the temperature rises. Are what temperature will the cylinder explode?

P1=12 ATM T1=27'C or 273+27=300 K P2=14.9 ATM therefore, P1/T1 = P2/T2 =>T2=(P2 x T1)/P1 =(14.9 x 300)/12 =372.5 K or 372.5 -273=99.5'C

LPG cylinder can withstand a pressure of 14.9 atm. The pressure gauge of the cylinder reads 12 atm at 27^°C. Due to a sudden fire in the building the temperature rises. Are what temperature will the cylinder explode?

P1=12 ATM

T1=27'C or 273+27=300 K

P2=14.9 ATM

therefore,
P1/T1 = P2/T2
=>T2=(P2 x T1)/P1

=(14.9 x 300)/12

=372.5 K or 372.5 -273=99.5'C