A closed container in lab contains a sample of oxygen gas. When the temperature in lab is 16.3 degrees Celcius, the pressure in the container is 9.30 atm. What would the temperature in lab be (in Celcius) if the pressure in the container was 17.7 atm? Assume the mol and volume of the container are constant. Record your answer as a number with the correct significant figures.

T₂  = 550.60 K

Explanation:

Given data:

Initial temperature = 16.3°C (16.3+273 = 289.3 K)

Initial pressure = 9.30 atm

Final temperature = ?

Final pressure = 17.7 atm

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:

9.30 atm / 289.3 K = 17.7 atm/T₂

T₂ =  17.7 atm × 289.3  K / 9.30 atm

T₂  = 5120.61 atm. K /9.30 atm

T₂  = 550.60 K

imma come back to you hold on

since the atomic number is the number of protons in the nucleus, as we move to the left on the periodic table, the electrostatic pull between the nucleus and the valence electron shell decreases. as we move down a group (vertical column), the valence shell energy level gets further and further from the nucleus.