Question

A cook puts 1.90 g of water in a 2.00 L pressure cooker that is then warmed from the kitchen temperature of 20°C to 111°C. What is the pressure (in atm) inside the container?

Answer 1

Answer:

P= 168258.30696 Pa

Explanation:

Given that

Mass of water vapor m = 19.00 g

Volume of water vapor V = 2.00 L

$=2.00\times10^{-3}m^3$

Temperature of water vapor is T = 111°C

                                                = 384K

Molar mass of water is M = 18.0148 g/mol

Number of moles are

n = m/M

= (1.90 g)/(18.0148 g/mol)

= 0.1054 mol

Pressure inside the container is

P= nRT/V

$P=\frac{(0.1054)(8.314472)(384)}{(2.00\times10^{-3})}$

P= 168258.30696 Pa