Question about thermodynamics, it's too long to add here so its in the description.

Question

Question:

At 100 °C, 1.00 gram of liquid water has a volume of 1.043 mL. What volume will 1.00 gram of water vapor occupy at 100 °C at a pressure of 1.00 bar? What is the percent volume change in the the sample?

Answer 1

Here we use the ideal gas law PV = nRT where:

P = 1 bar

n = 1g/18.02g/mol = 0.0554 mol water

R = 0.08314 L * bar/(mol * K)

T = 100 C = 373 K

V = nrt/P = (0.0554 mol)*(0.08314 L * bar/(mol*K))*(373 K)/(1 bar)

V = 1.721 L

The percent volume change is 1721 mL/1.043 mL = 1650