A cooking cylinder can withstand a pressure of 24 ATM at 27 degree Celsius the pressure of the containing gas is a 12 ATM than the minimum temperature above which cylinder will explode is 1)327⁰c
2)273⁰c
3)300k
4)600k

Option A. 327 °C

Explanation:

The following data were obtained from the question:

Initial pressure (P1) = 12 atm

Initial temperature (T1) = 27 °C

Final pressure (P2) = 24 atm

Final temperature (T2) =?

Next, we shall convert 27 °C to Kelvin temperature. This can be obtained as follow:

Temperature (K) = Temperature (°C) + 273

Initial temperature (T1) = 27 °C

Initial temperature (T1) = 27 °C + 273

Initial temperature (T1) = 300 K

Next, we shall the temperature at which the cylinder will explode as follow:

Initial pressure (P1) = 12 atm

Initial temperature (T1) = 300 K

Final pressure (P2) = 24 atm

Final temperature (T2) =?

P1/T1 = P2/T2

12/300 = 24/T2

Cross multiply

12 × T2 = 300 × 24

Divide both side by 12

T2 = (300 × 24) /12

T2 = 600 K

Finally, we shall convert 600 K to celsius temperature.

This can be obtained as follow:

Temperature (°C) = Temperature (K) – 273

Temperature (K) = 600 K

Temperature (°C) = 600 – 273

Temperature (°C) = 327 °C

Therefore, the minimum temperature at which the cylinder will explode is

327 °C.

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