Notifications
Clear all
States of Matter
1
Posts
2
Users
0
Likes
476
Views
0
09/09/2021 7:34 pm
Topic starter
At STP, a container has 1 mole of He, 2 mole Ne, 3 mole O2 and 4 mole N2 without changing total pressure if 2 mole of O2 is removed. The partial pressure of O2 will be decreased by:
(a) 26%
(b) 40%
(c) 58.33%
(d) 66.66%
1 Answer
0
09/09/2021 7:40 pm
Correct answer: (c) 58.33%
Explanation:
PO2 = \(\frac{3}{10}\) x PT
After removing 2 mole of O2,
P'O2 = \(\frac{1}{10}\) x PT
Decrease in partial pressure of O2
= \(\frac{\frac{3P_T}{10} - \frac{P_T}{8}}{\frac{3P_T}{10}}\) x 100
= 58.33%
This post was modified 3 years ago by Shivani siva