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

Ideal gas equation pv=nRT (or) `T=(pV)/(nR)`
p=3.32 bar V,=`5dm^3`
n=4.0 , R=0.083 bar `dm^3k^-1mol^-1`
Substituting these values ,
`p=(3.32 ba r times 5 dm^3)/(4mol times 0.083 ba rdm^3k^-1mol^-1)`
`=50k(or)-223^@C`