Answer to Question #162275 in General Chemistry for Ryuka

Question

A sample of oxygen gas occupies a volume of 0.4652 L. When the volume is increased to 0.5782 L, the new pressure is 341.3 atm. What was the original pressure? Gas Law used:_______________

Answer 1

n= pV/RT = 341.3 x 0.5782/8.314 x 300 = 0,079119421.

p=nRT/V= 0,079119421 x 8.314 x 300/0.4652 = 424,203912 atm.