Question

If 12.0 g of water is converted to steam in a 3.60 L pressure cooker held at a temperature of 108 °C, what pressure will be produced?

Answer 1

Well, we solve the Ideal Gas equation....and get `P~=6*atm`

Explanation:

And so .....`P=(nRT)/V=((12.0*g)/(18.01*g*mol^-1)xx0.0821*(L*atm)/(K*mol)xx381.1*K)/(3.60*L)`

`-=5.79*atm`...that ought to get your curry cooked.