Two identical containers A and B have frictionless pistons. They contain the same volume of an ideal gas at the same temperature. The mass of the gas in A is m_Aand that in B is m_B. The gas in each cylinder is now allowed to expand isothermally to double the initial volume. The change in the pressure in A and B, respectively, is $∆$ p and 1.5 $∆$ p. Then

see full answer

Start JEE / NEET / Foundation preparation at rupees 99/day !!

21% of IItians & 23% of AIIMS delhi doctors are from Sri Chaitanya institute !!

An Intiative by Sri Chaitanya

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

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

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

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

answer is C.

(Unlock A.I Detailed Solution for FREE)

Ready to Test Your Skills?

Check your Performance Today with our Free Mock Test used by Toppers!

Take Free Test

Detailed Solution

For the gas in container A

$\begin{array}{l} ΔP = {(P_{A})}_{final} - {(P_{A})}_{intial} = \frac{n_{A} RT}{2 V} - \frac{n_{A} RT}{V} \\ ΔP = - \frac{n_{A} RT}{2 V} . . . . . . . . (i) \end{array}$

For gas in container B

$\begin{array}{l} 1 .5 ΔP = {(P_{B})}_{final} - {(P_{B})}_{intial} = \frac{n_{B} RT}{2 V} - \frac{n_{B} RT}{V} \\ 1 .5 ΔP = - \frac{n_{B} RT}{2 V} . . . . . . . . . . . . (ii) \end{array}$

From Eqs. (i) and (ii), we get $n_{B} = 1 .5 n_{A}$

$\Rightarrow 2 n_{B} = 3 n_{A} or 2 m_{B} = 3 m_{A}$

Watch 3-min video & get full concept clarity

tricks from toppers of Infinity Learn

Result Producing online coaching for JEE/NEET of south India

Largest All India Test Series for JEE/NEET

Best Books for IIT JEE

Best Books for NEET

🔥 Start Your JEE/NEET Prep at Just ₹1999 / month - Limited Offer! Check Now!

Get Expert Academic Guidance – Connect with a Counselor Today!

Result Producing online coaching for JEE/NEET of south India

Largest All India Test Series for JEE/NEET

Best Books for IIT JEE

Best Books for NEET