Question

The container shown in figure has two chambers, separated by a partition, of volumes $V_1=2.0litre$ and $V_2=3.0litre$. The chambers contain ${\mu }_1=4.0$ and ${\mu }_2=5.0$ moles of a gas at pressures $p_1=1.0atm$ and $p_2=2.0atm$. Calculate the pressure after the partition is removed and the mixture attains equilibrium.

Answer 1

Given,

$V_1$ = 2.0 litre
$V_2$= 3.0 litre
${\mu }_1$ = 4.0 moles
${\mu }_2$ = 5.0 moles
$P_1$ = 1.00 atm
$P_2$ = 2.00 atm

From ideal gas equation:

$P_1{V}_1={\mu }_{1}R{T}_1$

$P_2{V}_2={\mu }_{2}R{T}_2$

Total number of moles remain same

$\mu ={\mu }_1+{\mu }_2$

Total volume will be $V=V_1+V_2$

Kinetic energy for $1$ mole $E=\frac{3}{2}PV$

$PV=\frac{2}{3}E$

For ${\mu }_1$ moles ${\mu }_1{P}_1{V}_1=\frac{2}{3}{\mu }_1{E}_1$

For ${\mu }_2$ moles ${\mu }_2{P}_2{V}_2=\frac{2}{3}{\mu }_2{E}_2$

Total energy is $({\mu }_1{E}_1+{\mu }_2{E}_2)$

$=\frac{3}{2}({\mu }_1{P}_1{V}_1+{\mu }_2{P}_2{V}_2)$

$PV=\frac{2}{3}{E}_{total}=\frac{2}{3}\mu E_{permole}$

Putting all the values

$P(V_1+V_2)=\frac{2}{3}\times \frac{3}{2}(P_1{V}_1+P_2{V}_2)$

$P=\frac{P_1{V}_1+P_2{V}_2}{V_1+V_2}$ ...$\left(i\right)$

$=\left(\frac{1.00\times 2.0+2.00\times 3.0}{2.0+3.0}\right)$atm

$=\frac{8.0}{5.0}=1.60atm$

Remarks: There is no loss of energy to ambient, that’s why this form of ideal gas law represented by equation $\left(i\right)$ becomes very useful for adiabatic changes.

$\text{Final Answer}:1.60atm$.