Answer:
The pressure inside the cooker is 1.0804 atm.
Explanation:
Boiling occurs when the vapor pressure becomes equal to atmospheric pressure.
For water, At standard conditions (Pressure = 1 atm) boiling occurs at 373.15 K.
So, Standard conditions:
T₁ = 373.15 K
P₁ = 1 atm
Given ,
The water boils at Temperature = 130 °C
The conversion of T( °C) to T(K) is shown below:
T(K) = T( °C) + 273.15
So, the temperature, T₂ = (130 + 273.15) K = 403.15 K
To find pressure inside the cooker (P₂) :
Applying Amontons's Law as:
[tex]\frac {P_1}{T_1}=\frac {P_2}{T_2}[/tex]
So,
[tex]P_2=\frac {P_1}{T_1} \times {T_2}[/tex]
[tex]P_2=\frac {1 atm}{373.15 K} \times {403.15 K}[/tex]
[tex]P_2=1.0804 atm[/tex]
Thus, The pressure inside the cooker is 1.0804 atm.
