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.

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 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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