If `10 g` of a gas at atmospheric pressue is cooled from `273^(@)C` to `0^(@)C`, keeping the volume constant, its pressure would become

Options

A. `1//2 atm`
B. `1//273 atm`
C. `2 atm`
D. `273 atm`

Answer

A

Explanation

`T_(1)= 273^(@)C= 273+273^(@)K= 546^(@)`
`T_(2)= 0^(@)C= 273+0^(@)C=273^(@)K`
According to Gay-Lussac's law
`(P_(1))/(T_(1))=(P_(2))/(T_(2)) :. P_(2)=(P_(1)T_(2))/(T_(1))=(1xx273^(@)K)/(546^(@)K) atm, (1)/(2) atm`