One litre of ideal gas is compressed isothermal at 0.72 m of Hg-column so that its volume becomes 0.9 litre. Find its stress, if the density of mercury is `13.6xx10^(3)kg//m^(3)`.

Explanation

Here `V_(1)=1` litre, `P_(1)=0.72m` of Hg column, `V_(2)=0.9` litre, `P_(2)=?`
As `P_(1)V_(1)=P_(2)V_(2)`
(or) `P_(2)=(P_(1)V_(1))/(V_2)=(0.72xx1)/(0.9)`
`=0.8m` of Hg column
`"Stress"="increase in pressure"=P_(2)-P_(1)`
`=0.8-0.72`
`=0.08m` of Hg column
`=0.08xx(13.6xx10^3)xx9.8N//m^(2)`
`=10.66xx10^(3)=1.07xx10^(4)N//m^(2)`