Question

The normal boiling point of ethanol is 78.4 °C and 101.3 KPa. The heat of vaporization for ethanol is 42.32 kJ/mol. Determine the vapor pressure of ethanol at 95.0 °C. Use the Clasius Clapeyron Equation: Ln [P2/P1] = - [∆Hvap /R] * [1/T2 - 1/T1]

Answer 1

169.4 kpa

Step-by-step explanation:

Okay, we are given the following parameters from the question and they are; normal boiling point of ethanol = 78.4 °C( 351.4 k) , pressure= 101.3 kpa = 101.3 × 10^3 pa, heat of vaporization for ethanol = 42.32 kJ/mol = 42.32 ×10^3 J/mol, vapor pressure of ethanol at 95.0 °C(368 k)= ??(not given). The question asked us to use the Clapeyron Equation for this particular Question,all we have to do is to slot in the values given and evaluate, soft !.

The Clapeyron Equation is given below;

==> ln [P2/P1] = - [∆Hvap /R] × [1/T2 - 1/T1].

ln [ P2/ 101.3 × 10^3] = - [42.32 × 10^3/ 8.314] × [ 1/368 - 1/ 351.4].

==> ln P2 - ln 101.3 × 10^3 =

==> ln P2 - 11.53 = - 5090.21 × [0.0027 - 0.0028).

==>ln P2 = 0.51 + 11.53.

==> ln P2 = 12.04.

P2 = e^12.04.

P2= 169396.940460334134 pa.

P2 = 169.4 kpa.

Answer 2

6.1×10^4Pa or 61KPa

Step-by-step explanation:

The Clausius-Clapeyron equation is used to estimate the vapour pressure at different temperature, once the enthalpy of vaporization and the vapor pressure at another temperature is given in the question. The detailed solution is shown in the image attached. The temperatures were converted to kelvin and the energy value was converted from kilojoule to joule since the value of the gas constant was given in unit of joule per mole per kelvin. The fact that lnx=2.303logx was also applied in the solution.