Q. A gas is heated in such a way so that its pressure and volume both becomes double. Again by lowering temperature, one fourth of initial number of moles of air has been taken in, to maintain the double volume and pressure. By what fraction, the temperature must have been raised finally?
A) $\frac{1}{5}\ times$
B) $\frac{4}{5}\ times$
C) $\frac{16}{5}times$
D) $\frac{8}{5}\ times$
Explanation
(c)
$p_{1} = p;V_{1} = V;p_{2} = 2p;V_{2} = 2V$
$\frac{p_{1}V_{1}}{T_{1}} = \frac{p_{2}V_{2}}{T_{2}}$
$\frac{pV}{T_{1}} = \frac{2p \times 2V}{T_{2}}$
$T_{2} = 4T_{1}$
When, air has been taken in and $p,\ V$ remain constant,
$n_{1} \bullet 4T_{1} = n_{2} \bullet T_{2}$
$n_{1} = n$
and $n_{2} = n + \frac{1}{4}n = \frac{5}{4}n$
$\therefore\ n \bullet 4T_{1} = \frac{5}{4}n \bullet T_{2}$
$T_{2} = \frac{16}{5}T_{1}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.