Two identical containers A and B with frictionless positions contain the same ideal gas at the same temperature and the same volume V. The mass of the gas in A is m_A and that of B is m_B. The gas in each cylinder is now allowed to expand isothermally to the same final volume 3V. The change in the pressure in A and B are found to be $∆ P$ and 2.5 $∆ P$ respectively. Then

see full answer

Your Exam Success, Personally Taken Care Of

1:1 expert mentors customize learning to your strength and weaknesses – so you score higher in school , IIT JEE and NEET entrance exams.

An Intiative by Sri Chaitanya

$3 m_{A} = 2 m_{B}$

$2 m_{A} = 3 m_{B}$

$5 m_{A} = 2 m_{B}$

$5 m_{A} = 3 m_{B}$

answer is C.

(Unlock A.I Detailed Solution for FREE)

Best Courses for You

JEE

NEET

Foundation JEE

Foundation NEET

CBSE

Detailed Solution

T = Constant, $Pα \frac{1}{V}$
For Chamber A, $ΔP = P_{i} - P_{t} = \frac{n_{A} RT}{V} - \frac{n_{A} RT}{3 V} = \frac{2 nART}{3 V}$ ------(I)
For Chamber B, $2 .5 ΔP = P_{i} - P_{t} = \frac{n_{B} RT}{V} - \frac{n_{B} RT}{3 V} = \frac{2}{3} \frac{n_{s} RT}{V}$ ------(II)
From equations (I) and (II)
$\begin{array}{l} \frac{n_{A}}{n_{s}} = \frac{1}{2 .5} = \frac{2}{5} \\ \Rightarrow \frac{m_{A} / M}{m_{B} / M} = \frac{2}{5} \Rightarrow 5 m_{A} = 2 m_{B} \end{array}$