Q. 7

# A gas cylinder has walls that can bear a maximum pressure of 1.0 × 10^{6} Pa. It contains a gas at 8.0 × 10^{5} Pa and 300 K. The cylinder is steadily heated. Neglecting any change in the volume, calculate the temperature at which the cylinder will break.

Answer :

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 P_{max}=P_{2}=1.0 × 10^{6} Pa

Pressure of gas P_{1}=8.0 × 10^{5} Pa

Temperature of gas T_{1}=300K

Since the volume has not been changed therefore,

V_{1}=V_{2}=V

Hence number of moles will also be same n_{1}=n_{2}=n

Temperature at which the cylinder will break=T_{2}

Since n_{1} =n_{2} =n

Therefore,

T_{2} =375K

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

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