At constant pressure the volume of a gas is Vat `0^(@)` C. At which temperature it will be 3V?

Options

A. `300^(@)`
B. `273^(@)`
C. `573^(@)`
D. `546^(@)`

Answer

D

Explanation

`V_(1) =V`
`V_(2) = 3 V `
`T_(1) = 0^(@) C ` = 273 K
`T_(2) = ? `
According to Charle.s law
`(V_(1))/(T_(1)) = (V_(2))/(T_(2))`
`(V)/(273) = (3V)/(T_(2))`
`VT_(2) = 3Vxx 273 `
`T_(2) = (3V xx 273)/(V) = 819 K - 273 = 546^(@) C`