Question

What is the pressure of water vapour inside the container of volume 8.21litres at 300K when 0.1 mole of water is present

Answer 1

Answer: The pressure of water vapor inside the vessel is 0.3 atm

Explanation:

The equation given by ideal gas follows:

$PV=nRT$

where,

P = pressure of water vapor = ?

V = Volume of container = 8.21 L

T = Temperature of container = 300 K

R = Gas constant = $0.0821\text{ L. atm }mol^{-1}K^{-1}$

n = number of moles of water vapor = 0.1 moles

Putting values in above equation, we get:

$P\times 8.21L=0.1mol\times 0.0821\text{ L atm }mol^{-1}K^{-1}\times 300K\\\\P=\frac{0.1\times 0.0821\times 300}{8.21}=0.3atm$

Hence, the pressure of water vapor inside the vessel is 0.3 atm