Calculate the temperature of 4.0 mol of a gas occupying `5dm^3` at 3.32 bar.
`(R=0.083 ba r dm^3 K^-1 mol^-1)`.

Explanation

formulae :-
` P= 3.32 ` bar
` V= 5 dm^3`
`R = 0.083 ` bar ` dm^3 k^(-1) mol^(-1)`
` n=4 ` moles
` PV = nRT`
`T = (PV )/(nR )`
`= ( 3.32 xx 5 )/( 4 xx 0.0 83 ) = ( 16. 6 )/( 0.332 ) = 50`
` therefore T= 50 K`