Answer :
Answer:
Enthalpy is 44.95 kJ/mol
Solution:
As per the question:
Temperature, T = 270.6 K
Temperature, T' = 287.5 K
Pressure, P = 324.5 mmHg
Pressure, P' = 626.9 mmHg
Now,
To calculate the enthalpy, we make use of the Clausius-Clapeyron eqn:
[tex]ln\frac{P}{P'} = \frac{\Delta H}{R}(\frac{1}{T'} - \frac{1}{T})[/tex]
where
[tex]\Delta H = Enthalpy[/tex]
R = Rydberg's constant
Substituting suitable values in the above eqn:
[tex]ln\frac{324.5}{626.9} = \frac{\Delta H}{8.31447}(\frac{1}{287.5} - \frac{1}{270.6})[/tex]
[tex]- 0.658 = \frac{\Delta H}{8.31447}(\frac{1}{287.5} - \frac{1}{270.6})[/tex]
[tex]\Delta H = 44.95\ kJ/mol[/tex]