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

Explanation

`PV=nRT`
`P=3.32 "bar", V=5 dm^(3)`
`n=4 mol`,
`R=0.083 "bar" dm^(3) K^(-1) mol^(-1)`
Therefore
`3.32xx5=4xx0.083xxT`
or `T=(3.32xx5)/(4xx0.083)=50 K`