1.0 g of benzene is burnt completely in presence of 4.0g `O_(2)` in a completely evacuated bomb calorimeter of volume 1L. What is the pressure inside the bomb at `30^(@)C`, if the volume and pressure of water vapour produced are neglected?

Explanation

1g of benzene`=(1)/(78)=0.0128` mol of benzene
4g of `O_(2)=(4)/(32)=0.125` mol of `O_(2)`
`underset("0.0128 mol"(15)/(2)xx0.0128=0.096mol)(C_(6)H_(6)(l)+(15)/(2)O_(2)(g)) to underset(6xx0.0128=0.0768mol)(6CO_(2)(g)+3H_(2)O(g))`
Total number of moles of `CO_(2)` produced and `O_(2)` remained=[0.0768+(0.125-0.096)] mol=0.1058 mol
`P=(nRT)/(V)=(0.1058xx0.0821xx303)/(1)=2.63` atm