Question

The instruction booklet for your pressure cooker indicates that its highest setting is 13.5 psi . You know that standard atmospheric pressure is 14.7 psi, so the booklet must mean 13.5 psi above atmospheric pressure. At what temperature in degrees Celsius will your food cook in this pressure cooker set on "high"?
Also how do I work the equation out with the ln?

Answer 1

Answer:

119°C

Explanation:

This question can be solved by using  the Clausius-Clayperon Equation.

The formula is given as:

$In(\frac{P_2}{P_1})$ = $(\frac{delta H_{vap}}{R} )$$(\frac{1}{T_1}-\frac{1}{T_2})$

We can all figured it out  that the Heat of vaporization (Δ$H_{vap}$)= 40.65 kJ/mol or 40,650 J/mol

Now, P₂ (the highest pressure setting) = 14.7 psi + 13.5 psi

=28.2 psi

Also the standard atmospheric pressure (P₁) of $H_2O$ is 14.7 psi (1 atm) and  373K $(T_1)$

∴ substituting our data into the above equation, we have

$In(\frac{28.2 psi}{14,7psi})$ = $(\frac{40650J/mol}{8.314J/mol*k} )(\frac{1}{373}-\frac{1}{T_2} )$

To work out the (In) part in the Left-hand side; we input that directly on the scientific calculator and we  have;

0.65 = (4890) $(\frac{1}{373} -\frac{1}{T_2})$

0.65 = $(\frac{1}{3}*4890)-(\frac{1}{T_2}*4890)$

0.65 = 13.1 - $\frac{4890}{T_2}$

0.65 - 13.1 = - $\frac{4890}{T_2}$

-12.45 = - $\frac{4890}{T_2}$

$T_2$ = $(\frac{-4890}{-12.45})$

$T_2$ = 392K

We are tasked with leaving our answer at degree Celsius, therefore 392k to Celsius will be (392-273)°C

= 119°C