At `25^(@)C` and 760 mm of Hg pressure a gas occupies 600 mL volume. What will be its pressure at a height where temperature is `10^(@)C `and volume of the gas is 640 mL.

Explanation

`p_(1)=760mm Hg.V_(1)=600 mL`
`T_(1)2=25+273=298K`
`V_(2)=640mL and T_(2)-10+273=283K`
According to Combined gas law
`(p_(1)V_(1))/(T_(1))=(p_(2)V_(2))/(T_(2))`
`rArrp_(2)=(p_(1)V_(1)T_(1))/(T_(2))`
`rArrp_(2)=((760mmHg)xx(600mL)xx(283K))/((640mL)xx(298K))`
= 676.6 mm Hg