Question

The heat of vaporization for benzaldehyde is 48.8 kJ/mol, and its normal boiling point is 451.0 K. Use this information to determine benzaldehyde’s vapor pressure (in torr) at 53.5°C. Report your answer to three significant digits.

Answer 1

`p_2=5.17torr`

Step-by-step explanation:

Hello.

In this case, by using the Clausius-Clapeyron Equation which allows us to relate the vapor pressure, temperature and heat of vaporization as shown below:

`ln((p_1)/(p_2) )=(\Delta _vH)/(R)((1)/(T_2)-(1)/(T_1) )`

Whereas
`p_1` is 760 torr due to the normal conditions. In such a way, for computing the vapor pressure of benzaldehyde at 53.5 °C (326.65 K), we proceed as shown below:

`(p_1)/(p_2) =exp[(48800J/mol)/(8.314(J)/(mol*K))((1)/(326.65K)-(1)/(451.0K) )]\\\\(p_1)/(p_2)=147.0`

Thus, the vapor pressure at the final T is:

`p_2=(p_1)/(147.0)=(760torr)/(147.0)\\ \\p_2=5.17torr`

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