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//273 atm`
B. `2 atm`
C. `(1)/(2)atm`
D. `5.05xx10^(4)Nm^(-2)`

Explanation

Hint: Given,
`T_(1)=273^(@)C` or `546.15 K`
`P_(1)= 1 atm`
`T_(2)=0^(@)C` or `273.15 K`
`P_(2)=?`
`:. (P_(1))/(T_(1))=(P_(2))/(T_(2))`
`:. P_(2)=(P_(1)xxT_(2))/(T_(1))=(1xx273.15)/(546.15)`
`=(1)/(273.1)atm` or `5.05xx10^(4)Nm^(-2)`