What number of moles of oxygen would exert a pressure of 10 atm at 320 K in an 8.2 dm\(^{3}\) cylinder?
[R=0.082 atm dm\(^{-3}\) mol\(^{-1}\) K\(^{-1}\)]
- 0.32
- 1.52
-
3.13
- 31.25
The correct answer is: C
Explanation
P = 10atm, T= 320K, V= 8.2dm\(^{3}\)
PV =nRT
n = P * V/R * T
10 * 8.2/0.082 * 320
= 82/26.24
n = 3.13