Question

A round bottom flask contains 300 ml. of the gas at 27°C and 70 cm of Hg. Find the final pressure when it is transferred in a conical flask of 75 ml. capacity at a temperature of 57°C ?

Answer 1

Dear User,
Applying ideal gas equation i.e.
PV = nRT
Here, the gas is being transferred from one flask to the other. This means that the number of moles of gas is constant.
For flask 1:
P₁V₁=nRT₁
or, $\frac{{\mathrm{P}}_1{\mathrm{V}}_1}{{\mathrm{T}}_1}=\mathrm{nR}...\left(1\right)$
Similarly, for flask 2
$\frac{{\mathrm{P}}_2{\mathrm{V}}_2}{{\mathrm{T}}_2}=\mathrm{nR}...\left(2\right)$
Equating both equations we get,
$$\begin{gathered} \frac{{\mathrm{P}}_1{\mathrm{V}}_1}{{\mathrm{T}}_1}=\frac{{\mathrm{P}}_2{\mathrm{V}}_2}{{\mathrm{T}}_2} \\ {\mathrm{P}}_1=70\mathrm{cm}\mathrm{of}\mathrm{Hg}=\frac{70}{76}=0.9210\mathrm{atm},{\mathrm{P}}_2=? \\ {\mathrm{V}}_1=300\mathrm{ml},{\mathrm{V}}_2=75\mathrm{ml} \\ {\mathrm{T}}_1=27+273=300\mathrm{K},{\mathrm{T}}_2=57+273=330\mathrm{K} \\ \mathrm{Now}, \\ \frac{0.9210\times 300}{300}=\frac{{\mathrm{P}}_2\times 75}{330} \\ \mathrm{or},{\mathrm{P}}_2=4.0526\mathrm{atm}\mathrm{or}307.9\mathrm{mm}\mathrm{of}\mathrm{Hg} \end{gathered}$$