Question

PLZ HELP DUE IN 10MIN!!!!!!!!!The temperature of the water vapor (H2O) inside a pressure cooker increased from 295 K to 395 K. As

the sample is heated, the pressure of the water vapor went from 21.0 kPa to _____.

Answer 1

p2=28.12KPa

Step-by-step explanation:

Given data:

Initial temperature = 295 K

Final temperature= 395 K

Initial pressure = 21.0 KPa

Final pressure = ?

Solution:

According to Gay-Lussac Law,

The pressure of given amount of a gas is directly proportional to its temperature at constant volume and number of moles.

Mathematical relationship:

P₁/T₁ = P₂/T₂

Now we will put the values in formula:

21.0 KPa / 295 K = P₂/395 K

P₂ = 21.0 KPa× 395 K / 295 K

P₂ = 8295 KPa. K /295 K

P₂ = 28.12 KPa