Question

The instruction booklet for your pressure cooker indicates that its highest setting is 12.8 psi . You know that standard atmospheric pressure is 14.7 psi, so the booklet must mean 12.8 psi above atmospheric pressure. At what temperature in degrees Celsius will your food cook in this pressure cooker set on "high"?

Answer 1

Answer:At $237.57^oC$ food will cook in the pressure cooker set on high.

Explanation:

Standard atmospheric pressure $P_1$= 14.7 Psi = 1.0001 atm (1 Psi=0.06804 atm)

Standard temperature$T_1$ = 273.15 K

Highest pressure offered by pressure cooker: 12.8 Psi = 0.8709 atm

Pressure ($P_2$) inside the pressure cooker on the highest setting at temperature$T_2$ :

$P_2$ =  1.0001 atm + 0.8709 atm = 1.8710 atm

According to Gay lussac law  :

$(pressure)\propto (temperature)$ (at constant Volume)

$\frac{P_1}{T_1}=\frac{P_2}{T_2}$

$T_2=\frac{P_2\times T_1}{P_1}=\frac{1.8710 atm\times 273.15 K}{1.0001 atm}=510.73 K= 237.57^oC (T(^oC)=T-273.15 K)$

At $237.57^oC$  food will cook in the pressure cooker set on high.

Answer 2

Answer: The temperature in the pressure cooker set on high is 237.71°C

Explanation:

STP conditions:

Pressure = 14.7 psi

Temperature = 273 K

To calculate the final temperature of the system, we use the equation given by Gay-Lussac Law. This law states that pressure of the gas is directly proportional to the temperature of the gas at constant pressure.

Mathematically,

$\frac{P_1}{T_1}=\frac{P_2}{T_2}$

where,

$P_1\text{ and }T_1$ are the initial pressure and temperature of the gas.

$P_2\text{ and }T_2$ are the final pressure and temperature of the gas.

We are given:

$P_1=14.7psi\\T_1=273K\\P_2=(14.7+12.8)psi=27.5psi\\T_2=?$

Putting values in above equation, we get:

$\frac{14.7psi}{273K}=\frac{27.5psi}{T_2}\\\\T_2=510.71K$

Converting the temperature from kelvins to degree Celsius, by using the conversion factor:

$T(K)=T(^oC)+273$

$510.71=T(^oC)+273\\T(^oC)=237.71^oC$

Hence, the temperature in the pressure cooker set on high is 237.71°C