Question

if 12 grams of water is converted to steam in a 3.6L pressure cooker held at a temperature of 108 what pressure would be produced___________.

Answer 1

The pressure produced would be 5.82atm

Step-by-step explanation:

From the ideal gas equation,

PV = nRT

P = nRT/V

n = mass of water/molecular weight of water = 12/18 = 0.67mol, V = 3.6L = 3.6×1000cm^3 = 3600cm^3, assuming the temperature is in degree Celsius, T = 108°C = 108 + 273K = 381K, R = 82.057cm^3.atm/gmol.K

P = 0.67×82.057×381/3600 = 5.82atm