A fixed mass of gas at 1 10^5 Pa occupies 6 L. At constant temperature it is changed to 2 L. The new pressure is:
- 3 10^5 Pa
- 1.5 10^5 Pa
- 6 10^5 Pa
- 33,300 Pa
Answer
A) 3 10^5 Pa
p1V1 = p2V2 Rightarrow p2 = 1 10^5 62 = 3 10^5 Pa. SLO P-11-C-10 · Application
Physics MCQ · More Kinetic Theory and First Law of Thermodynamics MCQs · Practise