Answer:
[tex]H=41.3kJmol^{-1}[/tex]
Explanation:
The equation relating the the enthalphy, pressure and temperature is expressed as
[tex]ln(\frac{P_{2}}{P_{1}} )=\frac{H}{R}(\frac{1}{T_{1}}-\frac{1}{T_{2}} ) \\[/tex]
Where P is the pressure, H is the enthalphy, and T is the temperature.
since the given values are
[tex]T_{1}=293k, \\T_{2}=336.5k\\P_{1}=5.95kPA\\p_{2}=53.3kPA\\and R=8.314J.K^{-1}mol_{-1}[/tex]
if we insert values, we arrive at
[tex]ln(\frac{53.3}{5.95} )=\frac{H}{8.314}(\frac{1}{293}-\frac{1}{336.5} )\\2.19=\frac{H}{8.314}(0.00044)\\H=(2.19*8.314)/0.00044\\H=41,268.8Jmol^{-1}\\H=41.3kJmol^{-1}[/tex]
