Question

Starting at atmospheric pressure, by how much must the pressure change in order to lower the boiling point of water by 1° C? You may assume that these changes are both small, so you only need to compute first derivatives. Remember that, under most conditions, the volume (per molecule) of liquid water is small compared to that of water vapor. Remember: Atmospheric pressure = 101300 Pa Boiling point at atmospheric pressure = 373.15 K Latent heat (LB) of boiling water = 2.256 × 106 J/kg Molecular weight (mw) of water = 0.018 kg/mol

Answer 1

Answer:

ΔP = -3556.36 Pa

Explanation:

Atmospheric pressure, P = 101300 Pa

Latent heat of boiling water, $L_{B} = 2.256 * 10^{6} J/kg$

Boiling point at atmospheric pressure, T = 373.15 K

ΔT = 1 K

The change in volume as the water boils to vapor will be

$\triangle V = V_{v} - V_{l}$

Since the volume of vapor ($V_{v}$) is far greater than the volume of liquid ($V_{l}$)

$\triangle V = V_{v}$

Using the ideal gas equation:

$PV_{v} = RT\\V_{v} = RT/P$

R = 8.314 J/mol-K

$V_{v} = \frac{8.314 * 373.15}{101300} \\V_{v} = 0.0306 m^{3}$

ΔV = 0.0306 m³

Molecular weight of water, MW = 0.018 kg/mol

Using the Clausius - Clapeyron equation:

$\triangle P = \frac{L_B \triangle T}{T \triangle V}* MW\\\triangle P = \frac{2.256 * 10^{6} *1}{373.15* 0.0306}* 0.018$

ΔP = -3556.36 Pa

But the pressure change is said to lower the boiling point of water, only a negative pressure change will lower the boiling point.

ΔP = 3556.36 Pa