Question

In the given figure container A holds an ideal gas at a pressure of $50\times 10^{5}\, Pa$ and a temperature of $300 \,K$. It is connected by a thin tube (and a closed) container $B$, with four times the volume of $A$. Container $B$ holds same at a pressure of $1.0\times 10^{5} \, Pa$ and a temperature of $400 \,K$. The valve is open allow the pressures to equalize, but the temperature of each container constant at its initial value. The final pressure in the two containers will be to

• A. $1.5\times 10^{5} \, Pa$
• B. $2.5\times 10^{5}\, Pa$
• C. $2.1\times 10^{5} \, Pa$
• D. $3.5\times 10^{5} \, Pa$

Answer 1

Answer: (A)

$1.5\times 10^{5} \, Pa$

Answer 2

The correct option is(A): $1.5\times 10^{5} \, Pa$.

From ideal gas law For container $A, n_{1}=\frac{p_{1} V_{1}}{R T_{1}}$
For container $B$, $n_{2}=\frac{p_{2} V_{2}}{R T_{2}}$
After opening the value, $x$ moles of gas stream from container $A$ to container $B$ such that both container equalize at pressure $p$.
Number of moles in container $A$ has changed to $n_{1}-x$,
i.e., $\left(n_{1}-x\right)=\frac{p \cdot V_{1}}{R \cdot T_{1}}$
$\therefore x=n_{1}=\frac{p \cdot V_{1}}{R \cdot T_{1}}=\frac{\left(p_{1}-p\right) \cdot V_{1}}{R \cdot T_{1}} ...$(i)
Number of moles in container $6$ has changed to $n_{2}+x$,
therefore $\left(n_{2}+x\right)=\frac{p_{2} \cdot V_{2}}{R \cdot T_{2}}$
$\therefore x=\frac{p \cdot V_{2}}{R \cdot T_{2}}-n_{2}=\frac{\left(p-p_{z}\right) V_{2}}{R \cdot T_{2}} \ldots . .$ (ii)
Equating Eqs (i) and (ii), we get
$\frac{\left(p_{1}-p\right) \cdot V_{1}}{R \cdot T_{1}}=\frac{\left(p-p_{2}\right) \cdot V_{2}}{R \cdot T_{2}}$
$\Rightarrow$ $\left(p_{1}-p\right)=\left(p-p_{2}\right) \cdot\left(\frac{V_{2}}{V_{1}}\right) \cdot\left(\frac{T_{1}}{T_{2}}\right)$
The pressure changes in the two containers are proportional
$\left(p_{1}-p\right)=\left(p-p_{2}\right) . K$
with $K=\left(\frac{V_{2}}{V_{1}}\right) \cdot\left(\frac{T_{1}}{T_{2}}\right)$
$=4\left(\frac{300}{400}\right)=3$
$p=\frac{p_{1}+p_{2} \cdot K}{1+K}=\frac{5 \times 10^{5}+1 \times 10^{5}}{1+3}$
$=\frac{6 \times 10^{5}}{4}=1.5 \times 10^{5} Pa$