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

The Correct Option is A

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 \(K=\left(\frac{V_{2}}{V_{1}}\right) \cdot\left(\frac{T_{1}}{T_{2}}\right)\)
\(=4\left(\frac{300}{400}\right)=3\)
\(p=\frac{p_{1}+p_{2} \cdot K}{1+K}=\frac{5 \times 10^{5}+1 \times 10^{5}}{1+3}\)
\(=\frac{6 \times 10^{5}}{4}=1.5 \times 10^{5} Pa\)