Question

A sample of water in the vapor phase (no liquid present) in a flask of constant volume exerts a pressure of 403 mmHg at 99°C. The flask is slowly cooled.

i) Assuming no condensation, use the Ideal Gas Law to calculate the pressure of the vapor at 91°C; at 75°C.

ii) Will condensation occur at 91°C? 75°C?

iii) On the basis of your answers in i) and ii), predict the pressure exerted by the water vapor in the flask at 91°C and 75°C.

Answer 1

A sample of water in the vapor phase (no liquid present) in a flask of constant volume exerts a pressure of 403 mmHg at 99°C. The flask is slowly cooled.

i) Assuming no condensation, use the Ideal Gas Law to calculate the pressure of the vapor at 91°C; at 75°C.

Solution :-

T1 = 99 C +273 = 372 K

P1 = 403 mmHg

T2 = 91 C +273 = 364 K

T3= 75 C + 273 = 348 K

Using the pressure and temperature relation lets calculate the pressure at the 91 C that is 364 K

P1/T1 = P2/T2

P2 = P1*T2 / T1

    = 403 mmHg * 364 K / 372 K

   = 394.3 mmHg

Now lets calculate the pressure at 75 C that is 348 K

P1/T1 = P2/T2

P2 = P1*T2 / T1

    = 403 mmHg * 348 K / 372 K

   = 377 mmHg

ii) Will condensation occur at 91°C? 75°C?

The condensation will occur at the givne temperatures.

iii) On the basis of your answers in i) and ii), predict the pressure exerted by the water vapor in the flask at 91°C and 75°C.

Solution :-

Enthalpy of vaporization of water is 40.7 kJ/mol that is 40700 J /mol

We know vapor pressure at 99 C that is 372 K = 403 mmHg

Now lets calculate the vapor pressure at 91 C that is 364 K

Ln(P2/P1)= DeltaH vap / R [(1/T1)-(1/T2)]

Ln(P2/403 mmHg) = 40700 J per mol / 8.314 J per mol K *[(1/372)-(1/364)]

Ln(P2/403 mmHg) = -0.28922

P2/403 mmHg = anti ln [-0.28922]

P2/403 mmHg = 0.749

P2 = 0.749 * 403 mmHg

P2 = 302 mmHg

So the vapor pressure at 91 C is 302 mmHg

Now lets calculate the vapor pressure at 75 C that is 348 K

Ln(P2/P1)= DeltaH vap / R [(1/T1)-(1/T2)]

Ln(P2/403 mmHg) = 40700 J per mol / 8.314 J per mol K *[(1/372)-(1/348)]

Ln(P2/403 mmHg) = -1.854*10^-4

P2/403 mmHg = anti ln [-1.854*10^-4]

P2/403 mmHg = 0.4035

P2 = 0.4035 * 403 mmHg

P2 = 162.6 mmHg

So the vapor pressure at 75 C is 162.6 mmHg