Question

A 2l vessel contains oxygen at a pressure of 380mm Hg at 27°c. 1.40g of N2 gas is introduced in the vessel. Will the pressure of the gaseous mixture increase or decrease and to what extent?

Answer 1

Dear Student,

$$\begin{gathered} \mathrm{We}\mathrm{know}\mathrm{that} \\ \mathrm{PV}=\mathrm{nRT} \\ {\mathrm{P}}_1=380\mathrm{mmHg}=\frac{380}{760}\mathrm{atm}=0.5\mathrm{atm} \\ \mathrm{V}=2\mathrm{L} \\ \mathrm{R}=0.0821{\mathrm{Latmmol}}^{-1}{\mathrm{K}}^{-1} \\ \mathrm{T}=27+273=300\mathrm{K} \\ \mathrm{Now},\mathrm{moles}\mathrm{of}\mathrm{oxygen}\mathrm{in}\mathrm{the}\mathrm{vessel},\mathrm{n}=\frac{{\mathrm{P}}_1\mathrm{V}}{\mathrm{RT}} \\ \mathrm{or},\mathrm{n}=\frac{0.5\times 2}{0.0821\times 300}=0.0406\mathrm{mol} \\ \mathrm{Moles}\mathrm{of}{\mathrm{N}}_2\mathrm{introduced}=\frac{\mathrm{mass}}{\mathrm{molar}\mathrm{mass}}=\frac{1.40}{28}=0.05\mathrm{mol} \\ \mathrm{So},\mathrm{total}\mathrm{moles}\mathrm{of}\mathrm{gas}\mathrm{in}\mathrm{the}\mathrm{vessel},\mathrm{n}\text{'}=0.05+0.0406=0.0906\mathrm{mol} \\ \mathrm{The}\mathrm{pressure}\mathrm{will}\mathrm{obviously}\mathrm{increases}\mathrm{as}\mathrm{it}\mathrm{is}\mathrm{directly}\mathrm{related}\mathrm{to}\mathrm{the}\mathrm{number}\mathrm{of}\mathrm{moles}. \\ \mathrm{Here},\mathrm{moles}\mathrm{is}\mathrm{increasing},\mathrm{so}\mathrm{pressure}\mathrm{will}\mathrm{also}\mathrm{increase}. \\ \mathrm{now}\mathrm{pressure}\mathrm{of}\mathrm{the}\mathrm{vessel},{\mathrm{P}}_2=\frac{\mathrm{n}\text{'}\mathrm{RT}}{\mathrm{V}}=\frac{0.0906\times 0.0821\times 300}{2}=1.1157\mathrm{atm} \\ \mathrm{Therefore},\mathrm{increase}\mathrm{in}\mathrm{pressure}={\mathrm{P}}_2-{\mathrm{P}}_1=1.1157-0.5\mathrm{atm}=0.6157\mathrm{atm}=0.6157\times 760\mathrm{mmHg}=378.88\mathrm{mmHg} \end{gathered}$$