What will be the pressure of 10 gram of a gas kept under atmospheric pressure, if its temperature is changed from 546 K to 273 K ?

What will be the pressure of 10 gram of a gas kept under atmospheric p

1.	What will be the pressure of 10 gram of a gas kept under atmospheric pressure, if its temperature is changed from 546 K to 273 K ?
Answer» `(1)/(2)` BAR 273 bar 2 bar `(1)/(273)` bar Solution :Initial temperature `= T_(1)=546` Kelvin Initial pressure = ATMOSPHERIC pressure `= P_(1)=1` bar Final temp. `= T_(2)=273` Kelvin. Final pressure `= P_(2)=` According to Gay Lucca.s Law : `P prop T` (n.V constant) So, the temperature will be 273 K from 546 K. Means if we half it, them the pressure will also be halved, it will be `(1)/(2)` bar from 1 bar. OR `(P_(1))/(T_(1))=(P_(2))/(T_(2))` (n.V constant) `therefore ("1 bar")/("546 kelvin")=(P_(2))/("273 kelvin")` `therefore P_(2)=("273 kelving" xx"1 bar")/("546 kelvin")=(1)/(2)` bar