27.4 l of oxygen gas at 25.0 degree 1.30 atm and 8.50 l of helium gas at 25.0 degrees and 2.00 atm were pumped into a tank with a volume of 5.81 at 25 degrees. Calculate the new partial pressure of oxygen 6.13 atm.

Expert Answers

An illustration of the letter 'A' in a speech bubbles

This question can be answered using Boyle's Law. Notice that initially the oxygen gas and the helium gas have the same temperature, and when they are both pumped into a tank, they remain at the same temperature. So, the temperature remains constant throughout the process, and Boyle's Law can be...

See
This Answer Now

Start your 48-hour free trial to unlock this answer and thousands more. Enjoy eNotes ad-free and cancel anytime.

Get 48 Hours Free Access

This question can be answered using Boyle's Law. Notice that initially the oxygen gas and the helium gas have the same temperature, and when they are both pumped into a tank, they remain at the same temperature. So, the temperature remains constant throughout the process, and Boyle's Law can be applied.

As oxygen is pumped into a tank with the given volume, according to the Boyle's Law, the product of the pressure and the volume remains constant:

From here, the final pressure of oxygen is

`P_f = (P_iV_i)/(V_f) =(1.3*27.4)/5.81 = 6.13 atm `

(I am assuming that the volume of the tank is also given in liters.)

The new partial pressure of oxygen is 6.13 atm.

Approved by eNotes Editorial Team