# Two flasks I and II shown below are connected by a valve of negligible volume.

Question:

Two flasks I and II shown below are connected by a valve of negligible volume.

When the valve is opened, the final pressure of the system in bar is $x \times 10^{-2}$. The value of $x$ is__________

[Assume-Ideal gas; 1 bar $=10^{5} \mathrm{~Pa}$; Molar mass of $\left.\mathrm{N}_{2}=28.0 \mathrm{~g} \mathrm{~mol}^{-1} ; \mathrm{R}=8.31 \mathrm{~J} \mathrm{~mol}^{-1} \mathrm{~K}^{-1}\right]$

Solution:

$\Rightarrow\left(\begin{array}{c}\text { Heat lost by } \\ \mathrm{N}_{2} \text { of container } \\ \mathrm{I}\end{array}\right)=\left(\begin{array}{c}\text { Heat gained by } \\ \mathrm{N}_{2} \text { of container } \\ \mathrm{II}\end{array}\right)$

$\Rightarrow \mathrm{n}_{1} \mathrm{C}_{\mathrm{m}}(300-\mathrm{T})=\mathrm{n}_{\mathrm{n}} \mathrm{C}_{\mathrm{m}}(\mathrm{T}-60)$

$\Rightarrow\left(\frac{2.8}{28}\right)(300-\mathrm{T})=\frac{0.2}{28}(\mathrm{~T}-60)$

$\Rightarrow 14(300-\mathrm{T})=\mathrm{T}-60$

$\Rightarrow \frac{(14 \times 300+60)}{15}=\mathrm{T}$

$\Rightarrow \mathrm{T}=284 \mathrm{~K}$ (final temperature)

$\Rightarrow$ If the final pressure $=\mathrm{P}$

$\Rightarrow\left(\mathrm{n}_{\mathrm{I}}+\mathrm{n}_{\mathrm{II}}\right)_{\text {final }}=\left(\frac{3.0}{28}\right)$

$\Rightarrow \frac{\mathrm{P}}{\mathrm{RT}}\left(\mathrm{V}_{\mathrm{I}}+\mathrm{V}_{\mathrm{II}}\right)=\frac{3.0 \mathrm{gm}}{28 \mathrm{gm} / \mathrm{mol}}$

$\mathrm{P}=\left(\frac{3}{28} \mathrm{~mol}\right) \times 8.31 \frac{\mathrm{J}}{\mathrm{mol}-\mathrm{K}} \times \frac{284 \mathrm{~K}}{3 \times 10^{-3} \mathrm{~m}^{3}} \times 10^{-5} \frac{\mathrm{bar}}{\mathrm{Pa}}$

$\Rightarrow 0.84287$ bar

$\Rightarrow 84.28 \times 10^{-2}$ bar

$\Rightarrow 84$

screw you
Sept. 25, 2023, 6:35 a.m.
other than copying other websites content, do something better....
screw you
Sept. 25, 2023, 6:35 a.m.
other than copying other websites content, do something better....