The vapor pressure of water at 25 degrees celsius is 23.8 torr, and the heat of vaporization of water at 25 degrees celsius is 43.9 kj/mol. calculate the vapor pressure of water at 50 degrees celsius.

The vapor pressure of water at 50 °C  is 93.7 torr.

Explanation:

The expression for Clausius-Clapeyron Equation is shown below as:

Where,  

P is the vapor pressure

ΔHvap  is the Enthalpy of Vaporization

R is the gas constant (8.314×10⁻³ kJ /mol K)

c is the constant.

For two situations and phases, the equation becomes:

Given:

= 23.8 torr

= ?

= 25°C

= 50 °C  

ΔHvap = 43.9 kJ/mol

The conversion of T( °C) to T(K) is shown below:

T(K) = T( °C) + 273.15  

So,  

T = (25 + 273.15) K = 298.15 K  

T = (50 + 273.15) K = 323.15 K  

= 298.15 K  

= 323.15 K

So,  applying in the above equation as:-

The vapor pressure of water at 50 °C  is 93.7 torr.

in the given question, the partial pressure is calculated by using the ideal gas equation, that is, pv = nrt

here, p is pressure, v is volume = 25 l, n is number of moles = 0.0104 moles, r is gas constant = -0.0821 l.atm / mol.k

t is temperature = 273 k

p = nrt / v

p = (0.0104 mol × 0.0821 l.atm / mol.k × 273 k) / 25 l

= 0.0093 atm

thats how its supposed to be