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.

1 answer

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.