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
Answers
Answer:
327 ° C
Step-by-step 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
Now we need to covert the kelvin into celsius
so 600 - 273
=327 ° C
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Answer:
27
is the answer ok !!!!!!!!!!!!!