Q. The temperature of an ideal at atmospheric pressure is $300K$ and volume $1m^{3}.$ If temperature and volume become double, then pressure will be
A) $10^{5}N/m^{2}$
B) $2 \times 10^{5}N/m^{2}$
C) $0.5 \times 10^{5}N/m^{2}$
D) $4 \times 10^{5}N/m^{2}$
Explanation
(a) $PV = \mu RT \Rightarrow P \propto \frac{T}{V}.$ If $T$ and $V$ both doubled then pressure remains same, $i.e.,\ P_{2} = P_{1} = 1\ atm = 1 \times 10^{5}N/m^{2}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.