A glass tube is inverted and dipped in mercury as shown. One mole of an ideal monoatomic gas is trapped in the tube and the tube is held so that length of the tube above the mercury level is always h₀ meter. The atmospheric pressure is equal to h₀ meter of mercury. The mercury vapour pressure, heat capacity of mercury, tube and the container are negligible. How much heat must be supplied to the gas inside the tube so as to increase its temperature by $∆$T ?

Let P₁ = initial pressure of the gas inside the tube
${\mathrm{P}}_0={\mathrm{P}}_1+{\mathrm{\rho gh}}_1$  [h₁ height of Hg column]
$\begin{array}{l}{\mathrm{\rho gh}}_0={\mathrm{P}}_1+{\mathrm{\rho gh}}_1\\ \therefore {\mathrm{P}}_1=\mathrm{\rho g}\left({\mathrm{h}}_0-{\mathrm{h}}_1\right)\end{array}$
Temperature of the gas is given by
$\begin{array}{l}{\mathrm{T}}_1=\frac{{\mathrm{P}}_1{\mathrm{V}}_1}{\mathrm{nR}}=\frac{\mathrm{\rho g}\left({\mathrm{h}}_0-{\mathrm{h}}_1\right)\mathrm{A}\left({\mathrm{h}}_0-{\mathrm{h}}_1\right)}{\mathrm{R}}[\mathrm{A}=\mathrm{area} \mathrm{of} \mathrm{cross} \mathrm{section}]\\ \therefore {\mathrm{T}}_1=\frac{{\left({\mathrm{h}}_0-{\mathrm{h}}_1\right)}^2\mathrm{\rho gA}}{\mathrm{R}}\end{array}$
After increasing the temperature, let the height of Hg column be h₂.
$\begin{array}{l}{\mathrm{T}}_2=\frac{{\left({\mathrm{h}}_0-{\mathrm{h}}_2\right)}^2\mathrm{\rho gA}}{\mathrm{R}}\\ \therefore \mathrm{\Delta T}={\mathrm{T}}_2-{\mathrm{T}}_1=\frac{\mathrm{\rho gA}}{\mathrm{R}}\left[{\left({\mathrm{h}}_0-{\mathrm{h}}_2\right)}^2-{\left({\mathrm{h}}_0-{\mathrm{h}}_1\right)}^2\right] .....\left(\mathrm{i}\right)\end{array}$
Let’s calculate the work done by the gas in pushing the Hg column.
$$\begin{gathered} \mathrm{W}=\int \mathrm{PdV}=-{\int }_{{\mathrm{h}}_1}^{{\mathrm{h}}_2} \mathrm{\rho g}\left({\mathrm{h}}_0-\mathrm{h}\right)\mathrm{Adh}[-\mathrm{ve} \mathrm{sign} \mathrm{as} \mathrm{dh} \mathrm{is} \mathrm{negative}] \\ \begin{array}{l}=\mathrm{\rho gA}{\left[\frac{{\left({\mathrm{h}}_0-\mathrm{h}\right)}^2}{2}\right]}_{{\mathrm{h}}_1}^{{\mathrm{h}}_2}\\ =\frac{\mathrm{\rho gA}}{2}\left[{\left({\mathrm{h}}_0-{\mathrm{h}}_2\right)}^2-{\left({\mathrm{h}}_0-{\mathrm{h}}_1\right)}^2\right]=\frac{\mathrm{R\Delta T}}{2}\left[\text{ using (i) }\right]\end{array} \end{gathered}$$
First law of thermodynamics
$\begin{array}{l}\mathrm{\Delta Q}={\mathrm{nC}}_{\mathrm{V}}\mathrm{\Delta T}+\mathrm{W}\\ =\frac{3}{2}\mathrm{R\Delta T}+\frac{\mathrm{R\Delta T}}{2}=2\mathrm{R\Delta T}\end{array}$