Q. 7

# A gas cylinder has walls that can bear a maximum pressure of 1.0 × 106 Pa. It contains a gas at 8.0 × 105 Pa and 300 K. The cylinder is steadily heated. Neglecting any change in the volume, calculate the temperature at which the cylinder will break.

We know ideal gas equation

PV=nRT

Where V= volume of gas

R=gas constant

T=temperature

n=number of moles of gas

P=pressure of gas.

Given

Maximum pressure Pmax=P2=1.0 × 106 Pa

Pressure of gas P1=8.0 × 105 Pa

Temperature of gas T1=300K

Since the volume has not been changed therefore,

V1=V2=V

Hence number of moles will also be same n1=n2=n

Temperature at which the cylinder will break=T2

Since n1 =n2 =n

Therefore,

T2 =375K

The temperature at which the cylinder will break=375K.

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