The graph drawn between pressure and volume in Boyle's law experiment is shown in figure for different masses of same gas at same temperature then

$m_{1} > m_{2}$
$m_{1}^{3} > m_{2}$
$m_{1} = m_{2}$
$m_{2} > m_{1}$

Question

The graph drawn between pressure and volume in Boyle-s law experiment is shown in figure different masses of same gas same temperature thenm-1- - m-2-m-1-3- - m-2-m-1- - m-2-m-2- - m-1-

Toppr Admin · Accepted Answer

From ideal gas equation- PV-nRTand n - wM-xA0- where M is the molecular weight of the gasNow PV - wMRTso- we draw a vertical line at a volume say V0-Then for this volume- we see that pressure is higher for the gas with molecular weight as M2 -from the graph- hence- this means that M2 must be smaller than M1

The graph drawn between pressure and volume in Boyle's law experiment is shown in figure for different masses of same gas at same temperature thenm1>m2m31>m2m1=m2m2>m1

The graph drawn between pressure and volume in Boyle's law experiment is shown in figure for different masses of same gas at same temperature then

$m_{1} > m_{2}$
$m_{1}^{3} > m_{2}$
$m_{1} = m_{2}$
$m_{2} > m_{1}$