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 ?

Options

A. `(1)/(2)` bar
B. 273 bar
C. 2 bar
D. `(1)/(273)` bar

Answer

A

Explanation

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