Two identical cylinders A and B with frictionless pistons contain the same ideal gas at the same temperature and the same voluem V. The mass of gas `Ais m_(A) and that of B is m_(B)` .The gs in eah cylinder is notw allowed to expand isothermally to the same final volume 2V . The change in the pressure in A and B are found ot be `DeltaP and 1.5 DeltaP` respecitvely Then

Options

A. `4m _A = 9 m _B`
B. `2m_A = 3m_B`
C. `3m_A = 2m_B`
D. `9m_A = 4m_B`

Answer

C

Explanation

Process is isothermal Therefore, T = constant. Volume is increasing, therefore, pressure will decrease `(P prop 1/V)`
In chamber `to`
`Delta p (rho_A)_i - (rho_A)_f = (n_A RT)/V - (n_A RT)/(2V) = (n_A RT)/(2V)" ...(i)"`
In chamber `to`
`1.5 Delta p= (rho_B)_i -(rho_B )_f = (n_B RT)/V - (n_B RT)/(2V) = (n_B RT)/(2V)" ...(ii)"`
From Egs. (i) and (ii), ` n_A/n_B = 1/(1.5) = 2/3`
`(m_A //M)/(m_B //M) = 2/3," "m_A/m_B = 2/3, " "3m_A = 3m_B`