The density of a gas at `27^(@)C` and `760 mm` pressure is `24`. Calculate the temperature at which it will be `18`, the pressure remaining constant.

Explanation

Given
`d_(1)=24`
`T_(1)=27^(@)C` or `300.15 K`
`P_(1)=760 mm` or `1 atm`
`d_(2)=18`
`T_(2)=?`
`P_(2)=760 mm` or `1 atm`
We know density `=(PM)/(RT)`
`:. (d_(1))/(d_(2))=(P_(1))/(T_(1))xx(T_(2))/(P_(2))=(T_(2))/(T_(1))` ( `': P_(1)=P_(2)=1 atm`)
` :. T_(2)=(d_(1))/(d_(2))xxT_(1)`
or `T_(2)=(24)/(18)xx300.15=400.2K` or `127.05^(@)C`