A $1.0 L$ closed vessel at $300 K$ contains $N X_{2}$ at $0.800 atm$ together with $0.10 g$ of liquid water. The vapour pressure of water at $300 K$ is $0.035 atm$ . Taking $R = 0.0821 L atm K^{- 1} mo l^{- 1}$ , the total pressure in the vessel at equilibrium in atm (three decimals) is

Numerical answer type, answer in atm, tolerance 0.002.

Answer: 0.835 atm

If all the water vaporised:

n = 0.10/18 = 0.00556 mol

P = \frac{0.00556 \times 0.0821 \times 300}{1.0} = 0.137 atm

, which exceeds the vapour pressure. So liquid water remains and

p_{H X_{2} O} = 0.035 atm

. Total

= 0.800 + 0.035 = 0.835 atm