A pressure cooker is used to cook food in a closed pot. By heating the contents of a pressure cooker at constant volume, the pressure increases. If the steam
inside the pressure cooker is initially at 100. °C and 1.00 atm, what is the final temperature (in °C) of the steam if the pressure is increased to 1.20 atm? Be
sure your answer has the correct number of significant figures.
To solve this problem, we can use the ideal gas law:
PV = nRT
Where:
P = pressure
V = volume (constant in this case)
n = number of moles
R = ideal gas constant
T = temperature
First, we need to convert the initial pressure to Pascals (1 atm = 101325 Pa) and the final pressure to Pascals:
Initial pressure: 1.00 atm = 1.01325 x 10^5 Pa
Final pressure: 1.20 atm = 1.2159 x 10^5 Pa
Now we can set up two versions of the ideal gas law equation for the initial and final states:
For the initial state:
(1.01325 x 10^5 Pa)V = n(8.314 J/mol*K)T1
For the final state:
(1.2159 x 10^5 Pa)V = n(8.314 J/mol*K)T2
Since the volume is constant, it cancels out in the equations. We can divide the two equations to get:
(1.01325 x 10^5 Pa) / (1.2159 x 10^5 Pa) = T1 / T2
T2 = T1 * (1.01325 x 10^5 Pa) / (1.2159 x 10^5 Pa)
Now plug in the initial temperature and pressures:
T2 = 100 °C * (1.01325 x 10^5 Pa) / (1.2159 x 10^5 Pa)
T2 = 83.09 °C
Therefore, the final temperature of the steam inside the pressure cooker at 1.20 atm is 83.09 °C.