Q: Two identical contains A and B with frictionless pistons contain the same ideal gas at the same temperature and the same volume V. the mass of the gas in A is mA and that in B is mB. The gas in each cylinder is now allowed to expand isothermally to the same final volume 2V. the change in the pressure in A and B are found to be ∆p and 1.5 ∆p respectively. Then

(a) 4 mA = 9 mB

(b)2 mA = 3 mB

(c) 3 mA = 2 mB

(d) 9 mA = 4 mB

Ans: (c)

Sol: As process is isothermal , hence T = constant (P V = constant)

In chamber A :

$\large \Delta p = (p_A)_i – (p_A)_f $

$\large \Delta p = \frac{n_A R T}{V} – \frac{n_A R T}{2V} = \frac{n_A R T}{2V} $

In chamber B :

$\large 1.5\Delta p = (p_B)_i – (p_B)_f $

$\large 1.5\Delta p = \frac{n_B R T}{V} – \frac{n_B R T}{2V} = \frac{n_B R T}{2V} $

On dividing ;

$\large \frac{n_A}{n_B} = \frac{1}{1.5} = \frac{2}{3}$

$\large \frac{m_A/M}{m_B/M} = \frac{2}{3}$

$\large \frac{m_A}{m_B} = \frac{2}{3}$

$\large 3 m_A = 2 m_B$