Two identical containers A and B with frictio | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(C) The initial pressure in the two containers is given by the ideal gas law $PV = nRT = (m/M)RT$,where $M$ is the molar mass.$P_A = \frac{m_A RT}{MV}

Vedclass · Accepted Answer

$3m_A = 2m_B$

Vedclass · Answer

(C) The initial pressure in the two containers is given by the ideal gas law $PV = nRT = (m/M)RT$,where $M$ is the molar mass.$P_A = \frac{m_A RT}{MV}$ and $P_B = \frac{m_B RT}{MV}$.After isothermal expansion to volume $2V$,the final pressures are:$P'_A = \frac{m_A RT}{M(2V)}$ and $P'_B = \frac{m_B RT}{M(2V)}$.The change in pressure for each container is:$\Delta P_A = P_A - P'_A = \frac{m_A RT}{MV} - \frac{m_A RT}{2MV} = \frac{m_A RT}{2MV}$.$\Delta P_B = P_B - P'_B = \frac{m_B RT}{MV} - \frac{m_B RT}{2MV} = \frac{m_B RT}{2MV}$.Given that $\Delta P_A = \Delta P$ and $\Delta P_B = 1.5 \Delta P$,we take the ratio:$\frac{\Delta P_A}{\Delta P_B} = \frac{\Delta P}{1.5 \Delta P} = \frac{1}{1.5} = \frac{2}{3}$.Substituting the expressions for $\Delta P_A$ and $\Delta P_B$:$\frac{m_A RT / 2MV}{m_B RT / 2MV} = \frac{2}{3} \implies \frac{m_A}{m_B} = \frac{2}{3}$.Therefore,$3m_A = 2m_B$.

Similar Questions

In the following arrangement,$y = 1.0 \, mm$,$d = 0.24 \, mm$,and $D = 1.2 \, m$. The work function of the material of the emitter is $2.2 \, eV$. The stopping potential $V$ needed to stop the photocurrent will be....$V$

$A$ particle moves in a straight line with retardation proportional to its displacement. Its loss of kinetic energy for any displacement $x$ is proportional to:

The value of $\frac{3 + \cot 76^\circ \cot 16^\circ}{\cot 76^\circ + \cot 16^\circ}$ is:

$Ge$ and $Si$ diodes conduct at $0.3 \ V$ and $0.7 \ V$ respectively. In the following figure,if the $Ge$ and $Si$ diode connections are reversed,the value of $V_0$ changes by .....$V$.

What is a species?

Vedclass Test Series

Exam Paper Generator

Online Exam Module