Question

Pure liquid water has a heat of vaporization of ­x kJ/mol. The vapor pressure of water is 23.756 torr at room temperature (25 °C). What would the vapor pressure of water be expected to be at 37 °C? How would the vapor pressure be changed if 1.0 gram of NaCl were added to a volume of 100 mL of water?

Answer 1

We have to use Clausius-Clapeyron Equation i.e.: ln (P₁/P₂) = (-ΔH_vap/R)(1/T₁ - 1/T₂) .....(i)

where, P₁ and P₂ are the respective pressures at temperatures T₁ and T₂.

ΔH_vap= is the heat of vaporization= x kJ/mol

So, P₁=23.756 torr ; P₂=? ; T₁= 25 °C= (273+25) K=298 K ; T₂= 37 °C= (273+37) K=310 K R= 8.314 JK⁻¹mol⁻¹

Therefore, using the above values in equation (i): ln (P₁/P₂) = (-ΔH_vap/R)(1/T₁ - 1/T₂)

ln (23.756/P₂) = (-X/R)(1/298 - 1/310)

ln 23.756 - ln P₂ = (-x/R)(1/298 - 1/310)

ln P₂ -ln 23.756 = (x/R)(1/298 - 1/310)

  ln P₂ = (x/R)(1/298 - 1/310) +ln 23.756

P₂ = e^{[(x/R)(1/298 - 1/310) +ln 23.756]}

If we put x=40.66 kJ/mol (which is the heat of vaporization of pure water at room temperature)

we get: P₂ = e^{[(40.66/R)(1/298 - 1/310) +ln 23.756]}

P₂ = 23.771 torr

So, vapor pressure of water at 37 °C = 23.771 torr

According to Raoult's Law, vapor pressure (P) of a solvent in a solution is equal to the vapour pressure of the pure solvent (P^∘) times its mole fraction.

So, P=(mole fraction of water)P^∘.....(ii)

mole fraction of water = (moles of water)/ (moles of NaCl + moles of H₂O)

moles of NaCl= wt/molecular weight = 1/58.44 =0.017 mol

moles of H₂O= wt/molecular weight = 100/18.02= 5.549 mol

So, mole fraction of water = 5.549/(0.017+5.549)= 0.9969

Using the above value in (ii) we get:

vapor pressure when 1g of NaCl is added to 100ml water at 25 °C : P=(0.9969)23.756=23.68 torr

vapor pressure when 1g of NaCl is added to 100ml water at 37 °C : P=(0.9969)23.771=23.69 torr.