Two identical containers A and B have frictionless pistons. They contain same volume of an ideal gas at the same temperature. The mass of the gas in A is $m_{A}$ and that in B is $m_{B} .$
The gas in each cylinder is allowed to expand isothermally to double the initial volume.The isothermally to double the initial volume. The changes in the pressure in A and B respectively are $Δ p$ and 1.5 $Δ P$ 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

$4 m_{A} = 9 m_{B}$

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

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

$9 m_{A} = 4 m_{B}$

answer is C.

(Unlock A.I Detailed Solution for FREE)

Best Courses for You

JEE

NEET

Foundation JEE

Foundation NEET

CBSE

Detailed Solution

$Δ P = {(P_{A})}_{f i n a l} - {(P_{A})}_{i n i t i a l} = \frac{N_{A} R T}{2 v} - \frac{N_{A} R T}{v} = - \frac{N_{A} R T}{2 v} 1.5 Δ P = {(P_{B})}_{f i n a l} - {(P_{B})}_{i n i t i a l} = \frac{N_{B} R T}{2 v} - \frac{N_{B} R T}{v} = - \frac{N_{B} R T}{2 v} \frac{1.5 Δ P}{Δ P} = \frac{N_{B} R T / 2 v}{N_{A} R T / 2 v} = \frac{n_{B}}{n_{A}} = \frac{m_{B}}{m_{A}} 1.5 = \frac{m_{B}}{m_{A}} = \frac{3}{2} = \frac{m_{B}}{m_{A}} 3 m_{A} = 2 m_{B}$