An LPG cylinder contains gas at a pressure of 300 kPa at 27°C. The cylinder can withstand the pressure of 1.2*10? Pa. The room in which the cylinder is kept catches fire. The minimum temperature at which the bursting of cylinder will take place is __________ °C. (Nearest integer)

3 Views|Posted 7 months ago
Asked by Shiksha User
1 Answer
A
7 months ago

P? = 300 kPa = 3 * 10? Pa = 3 bar
T? = 300 K
P? = 1.2 * 10? Pa = 12 * 10? Pa = 12 bar
T? =?
P? /T? = P? /T?
3/300 = 12/T?
T? = (12 * 300)/3 K = 1200 K
∴ T in °C = 1200 - 273 = 927°C

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