Question

Two glass bulbs of equal volume are connected by a narrow tube and are filled with a gas at 0°C at a pressure of 76 cm of mercury. One of the bulbs is then placed in melting ice and the other is placed in a water bath maintained at 62°C. What is the new value of the pressure inside the bulbs? The volume of the connecting tube is negligible.

Answer 1

$\begin{array}{l}\text{Here,}\\ {\mathit{\text{P}}}_1=0.76\text{m Hg}\\ {\mathit{\text{P}}}_2=P\\ T_1=273\text{K}\\ T_2=335K\\ {\text{Let each of the bulbs have n}}_1\text{moles initially.}\\ {\text{Let the number of moles left in second bulb after its pressure reached P be n}}_2\text{.}\\ \text{Applying equation of state, we get}\\ \frac{P_{1}V}{n_1{T}_1}=\frac{PV}{n_2{T}_2}\\ \Rightarrow \frac{0.76}{273n_1}=\frac{P}{335n_2}\\ \Rightarrow n_2=\frac{273P}{335\times 0.76}{n}_1\\ \text{Number of moles left in the second bulb after the temperature rose =}{n}_1-n_2\\ \\ =n_1-\frac{273P}{335\times 0.76}{n}_1\\ {\text{Let n}}_3\text{moles be left when pressure reached P.}\mathrm{Applying\; equation\; of\; state\; in\; the\; first\; bulb,}\mathrm{we}\mathrm{get}\\ \frac{P_{1}V}{n_1{T}_1}=\frac{PV}{n_3{T}_1}\\ \Rightarrow \frac{0.76}{n_1}=\frac{P}{n_3}\\ \Rightarrow n_3=\frac{P{n}_1}{0.76}\\ {\text{n}}_3=\text{}{\text{its own n}}_1\text{moles + the it recieved from the first}\\ n_3=n_1+(n_1-n_2)\\ \Rightarrow \frac{P{n}_1}{0.76}=n_1+n_1-\frac{273P}{335\times 0.76}{n}_1\\ \Rightarrow \frac{P}{0.76}=2-\frac{273P}{335\times 0.76}\\ \Rightarrow P=0.8375\\ \Rightarrow P=84\text{cm of Hg}\end{array}$